# Euclid, Book I, as translated by Thomas L. Heath

## Definitions [Heath's Edition]

1. A point is that which has no part.
2. A line is breadthless length.
3. The extremities of a line are points.
4. A straight line is a line which lies evenly with the points on itself.
5. A surface is that which has length and breadth only.
6. The extremities of a surface are lines.
7. A plane surface is a surface which lies evenly with the straight lines on itself.
8. A plane angle is the inclination to one another of two lines in a plane which meet one another and do not lie in a straight line.
9. And when the lines containing the angle are straight, the angle is called rectilineal.
10. When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.
11. An obtuse angle is an angle greater than a right angle.
12. An acute angle is an angle less than a right angle.
13. A boundary is that which is an extremity of anything.
14. A figure is that which is contained by any boundary or boundaries.
15. A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure are equal to one another;
16. And the point is called the centre of the circle.
17. A diameter of the circle is any straight line drawn through the centre and terminated in both directions by the circumference of the circle, and such a straight line also bisects the circle.
18. A semicircle is the figure contained by the diameter and the circumference cut off by it. And the centre of the semicircle is the same as that of the circle.
19. Rectilineal figures are those which are contained by straight lines, trilateral figures being those contained by three, quadrilateral those contained by four, and multilateral those contained by more than four straight lines.
20. Of trilateral figures, an equilateral triangle is that which has its three sides equal, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle that which has its three sides unequal.
21. Further, of trilateral figures, a right-angled triangle is that which has a right angle, an obtuse-angled triangle that which has an obtuse angle, and an acute-angled triangle that which has its three angles acute.
22. Of quadrilateral figures, a square is that which is both equilateral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right angled. And let quadrilaterals other than these be called trapezia.
23. Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction.

## Definitions [Lardner's Edition]

 (1) I. A point is that which has no parts. (2) II. A line is length without breadth. (3) III. The extremities of a line are points. (4) IV. A right line is that which lies evenly between its extremities. (5) V. A surface is that which has length and breadth only. (6) VI. The extremities of a surface are lines. (7) VII. A plane surface is that which lies evenly between its extremities. (9) VIII. A plane angle is the inclination of two lines to one another, in a plane, which meet together, but are not in the same direction. (10) IX. A plane rectilinear angle is the inclination of two right lines to one another, which meet together, but are not in the same right line. (11) X. When a right line standing on another right line makes the adjacent angles equal, each of these angles is called a right angle, and each of these lines is said to perpendicular to the other. (12) XI. An obtuse angle is an angle greater than a right angle. (13) XII. An acute angle is an angle less than a right angle. (15) XIII. A term or boundary is the extremity of any thing. (16) XIV. A figure is a surface, inclosed on all sides by a line or lines. (17) XV. A circle is a plane figure, bounded by one continued line, called its circumference or periphery; and having a certain point within it, from which all right lines drawn to its circumference are equal. (18) XVI. This point (from which the equal lines are drawn) is called the centre of the circle. (20) XVII. A diameter of a circle is a right line drawn through the centre, terminated both ways in the circumference. (21) XVIII. A semicircle is the figure contained by the diameter, and the part of the circle cut off by the diameter. (23) XIX. A segment of a circle is a figure contained by a right line, and the part of the circumference which it cuts off. (24) XX. A figure contained by right lines only, is called a rectilineal figure. (25) XXI. A triangle is a rectilinear figure included by three sides. (26) XXII. A quadrilateral figure is one which is bounded by four sides. The right lines A C, B D, connecting the vertices of the opposite sides of a quadrilateral figure, are called its diagonals. A B C D (27) XXIII. A polygon is a rectilinear figure, bounded by more than four sides. (28) XXIV. A triangle, whose three sides are equal, is said to be equilateral. (29) XXV. A triangle which has only two sides equal is called an isosceles triangle. (30) XXVI. A scalene triangle is one which has no two sides equal. (31) XXVII. A right-angled triangle is that which has a right angle. (32) XXVIII. An obtuse-angled triangle is that which has an obtuse angle. (33) XXIX. An acute-angled triangle is that which has an three acute angles. (34) XXX. An equilateral quadrilateral figure is called a lozenge. (35) XXXI. An equilateral lozenge is called a square. (35) XXXII. An oblong is a quadrilateral, whose angles are all right, but whose sides are not equal. (36) XXXIII. A rhomboid is a quadrilateral, whose opposite sides are equal. (37) XXXIV. All other quadrilateral figures are called trapeziums. (38) XXXV. Parallel right lines are such as are in the same plane, and which, being produced continually in both direction, would never meet.

## Postulates [Heath's Edition]

Let the following be postulated:

1. To draw a straight line from any point to any point.
2. To produce a finite straight line continuously in a straight line.
3. To describe a circle with any centre and distance.
4. That all right angles are equal to one another.
5. That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than two right angles.

## Postulates [Lardner's Edition]

 (39) I. Let it be granted that a right line may be drawn from any one point to any other point. (40) II. Let it be granted that a finite right line may be produced to any length in a right line. (41) III. Let it be granted that a circle may be described with any centre at any distance from that centre.

## Common Notions [Heath's Edition]

1. Things which are equal to the same thing are also equal to one another.
2. If equals be added to equals, the wholes are equal.
3. If equals be subtracted from equals, the remainders are equal.
4. Things which coincide with one another are equal to one another.
5. The whole is greater than the part.

## Axioms [Lardner's Edition]

 (43) I. Magnitudes which are equal to the same are equal to each other. (44) II. If equals be added to equals the sums will be equal. (45) III. If equals be taken away from equals the remainders will be equal. (46) IV. If equals be added to unequals the sums will be unequal. (47) V. If equals be taken away from unequals the remainders will be unequal. (48) VI. The doubles of the same or equal magnitudes are equal. (49) VII. The halves of the same or equal magnitudes are equal. (50) VIII. Magnitudes which coincide with one another, or exactly fill the same space, are equal. (51) IX. The whole is greater than its part. (52) X. Two right lines cannot include a space. (53) XI. All right angles are equal. (54) XII. A B C D If two right lines (A B, C D) meet a third right line (A C) so as to make the two interior angles (B A C and D C A) on the same side less than two right angles, these two right lines will meet if they be produced on that side on which the angles are less than two right angles.

## Euclid, Elements of Geometry, Book I, Proposition 1

### Proposition 1[Heath's Edition]

On a given finite straight line to construct an equilateral triangle.

Let AB be the given finite straight line.

Thus it is required to construct an equilateral triangle on the straight line AB.

With centre A and distance AB let the circle BCD be described; [Post. 3]
again, with centre B and distance BA let the circle ACE be described; [Post. 3]
and from the point C, in which the circles cut one another, to the points A, B let the straight lines CA, CB be joined. [Post. 1]

Now since the point A is the centre of the circle CDB,
AC is equal to AB. [Def. 15]

Again, since the point B is the centre of the circle CAE, BC is equal to BA. [Def. 15]

But CA was also proved equal to AB;
therefore each of the straight lines CA, CB is equal to AB.

And things which are equal to the same thing are also equal to one another; [C.N. 1]
therefore CA is also equal to CB.

Therefore the three straight lines CA, AB, BC are equal to one another.

Therefore the triangle ABC is equilateral; and it has been constructed on the given finite straight line AB.
(Being) what it was required to do.

Proposition I. Problem. [Lardner's Edition]

 (57) On a given finite right line (A B) to construct an equilateral triangle.

### Solution.

With the center A and the radius A B let a circle B C D be described (41), and with the centre B and the radius B A let another circle A C E be described. From a point of intersection C of these circles let right lines be drawn to the extremities A and B of the given right line (39). The triangle A C B will be that which is required.

### Demonstration.

It is evident that the triangle A C B is constructed on the given right line A B. But it is also equilateral; for the lines A C and A B, being radii of the same circle B C D, are equal (17), and also B C and B A, being radii of the same circle A C E, are equal. Hence the lines B C and A C, being equal to the same line A B, are equal to each other (43). The three sides of the triangle A B C are therefore equal, and it is an equilateral triangle (28).

(58)   In the solution of this problem it is assumed that the two circles intersect, inasmuch as the vertex of the equilateral triangle is a point of intersection. This, however, is sufficiently evident if it be considered that a circle is a continued line which includes space, and that in the present instance each circle passing through the centre of the other must have a part of its circumference within that other, and a part without it, and must therefore intersect it.

It follows from the solution, that as many different equilateral triangles can be constructed on the same right line as there are points in which the two circles intersect. It will hereafter be proved that two circles cannot intersect in more than two points, but for the present this may be taken for granted.

Since there are but two points of intersection of the circles, there can be but two equilateral triangles constructed on the same finite right line, and these are placed on opposite sides of it, their vertices being at the points C and F.

After having read the first book of the elements, the student will find no difficulty in proving that the triangles C F E and C D F are equilateral. These lines are not in the diagram, but may easily be supplied.

## Euclid, Elements of Geometry, Book I, Proposition 2

### Proposition 2[Heath's Edition]

To place at a given point (as an extremity) a straight line equal to a given straight line.

Let A be the given point, and BC the given straight line.

Thus it is required to place at the point A (as an extremity) a straight line equal to the given straight line BC.

From the point A to the point B let the straight line AB be joined; [Post. 1]
and on it let the equilateral triangle DAB be constructed. [I. 1]

Let the straight lines AE, BF be produced in a straight line with DA, DB; [Post. 2]
with centre B and distance BC let the circle CGH be described;
and again, with centre D and distance DG let the circle GKL be described. [Post. 3]

Then, since the point B is the centre of the circle CGH, BC is equal to BG.

Again, since the point D is the centre of the circle GKL, DL is equal to DG.

And in these DA is equal to DB;
therefore the remainder AL is equal to the remainder BG. [C.N. 3]

But BC was also proved equal to BG;
therefore each of the straight lines AL, BC is equal to BG.

And things which are equal to the same thing are also equal to one another; [C.N. 1]
therefore AL is also equal to BC.

Therefore at the given point A the straight line AL is placed equal to the given straight line BC.
(Being) what it was required to do.

Proposition II. Problem. [Lardner's Edition]

 (59) From a given point (A) to draw a right line equal to a given finite right line (B C).

### Solution.

Let a right line be drawn from the given point A to either extremity B of the given finite right line B C (39). On the line A B let an equilateral triangle A D B be constructed (I). With the centre B and the radius B C let a circle be described (41). Let D B be produced to meet the circumference of this circle in F (40), and with the centre D and the radius D F let another circle F L K be described. Let the line D A be produced to meet the circumference of this circle in L. The line A L is then the required line.

### Demonstration.

The lines D L and D F are equal, being radii of the same circle F L K (17). Also the lines D A and D B are equal, being sides of the equilateral triangle B D A. Taking the latter from the former, the remainders A L and B F are equal (45). But B F and B C are equal, being radii of the same circle F C H (17), and since A L and B C are both equal to B F, they are equal to each other (43). Hence A L is equal to B C, and is drawn from the given point A, and therefore solves the problem.

The different positions which the given right line and given point may have with respect to each other, are apt to occasion such changes in the diagram as to lead the student into error in the execution of the construction for the solution of this problem.

Hence it is necessary that in solving this problem, the student should be guided by certain general directions, which are independent of any particular arrangement which the several lines concerned in the solution may assume. If the student is governed by the following general directions, no change which the diagram can undergo will mislead him.

1° The given point is to be joined with either extremity of the given right line. (Let us call the extremity with which it is connected, the connected extremity of the given right line; and the line so connecting them, the joining line.)

2° The centre of the first circle is the connected extremity of the given right line; and its radius, the given right line.

3° The equilateral triangle may be constructed on either side of the joining line.

4° The side of the equilateral triangle which is produced to meet the circle, is that side which is opposite to the given point, and it is produced through the centre of the first circle till it meets its circumference.

5° The centre of the second circle is that vertex of the triangle which is opposite to the joining line, and its radius is made up of that side of the triangle which is opposite to the given point, and its production which is the radius of the first circle. So that the radius of the second circle is the sum of the side of the triangle and the radius of the first circle.

6° The side of the equilateral triangle which is produced through the given point to meet the second circle, is that side which is opposite to the connected extremity of the given right line, and the production of this side is the line which solves the problem; for the sum of this line and the side of the triangle is the radius of the second circle, but also the sum of the given right line (which is the radius of the first circle) and the side of the triangle is equal to the radius of the second circle. The side of the triangle being taken away the remainders are equal.

As the given point may be joined with either extremity, there may be two different joining lines, and as the triangle may be constructed on either side of each of these, there may be four different triangles; so the right line and the point being given, there are four different constructions by which the problem may be solved.

If the student inquires further, he will perceive that the solution may be effected also by producing the side of the triangle opposite the given point, not through the extremity of the right line but through the vertex of the triangle. The various consequences of this variety in the construction we leave to the student to trace.

(60)   By the second proposition a right line of a given length can be inflected from a given point P upon any given line A B. For from the point P draw a right line of the given length (II), and with P as centre, and that line as radius, describe a circle. A line drawn from P to any point C, where this circle meets the given line A B, will solve the problem.

By this proposition the first may be generalized; for an isosceles triangle may be constructed on a given line as base, and having its side of a given length. The construction will remain unaltered, except that the radius of each of the circles will be equal to the length of the side of the proposed triangle. If this length be not greater than half the base, the two circles will not intersect, and no triangle can be constructed, as will appear hereafter.

## Euclid, Elements of Geometry, Book I, Proposition 3

### Proposition 3[Heath's Edition]

Given two unequal straight lines, to cut off from the greater a straight line equal to the less.

Let AB, C be the two given unequal straight lines, and let AB be the greater of them.

Thus it is required to cut off from AB the greater a straight line equal to C the less.

At the point A let AD be placed equal to the straight line C; [I. 2]
and with centre A and distance AD let the circle DEF be described. [Post. 3]

Now, since the point A is the centre of the circle DEF,
AE is equal to AD. [Def. 15]

But C is also equal to AD.

Therefore each of the straight lines AE, C is equal to AD; so that AE is also equal to C. [C.N. 1]

Therefore, given the two straight lines AB, C, from AB the greater AE has been cut off equal to C the less.
(Being) what it was required to do.

Proposition III. Problem. [Lardner's Edition]

 (61) From the greater (A B), of two given right lines to cut off a part equal to the less (C).

### Solution.

From either extremity A of the greater let a right line A D be drawn equal to the less C (II), and with the point A as centre, and the radius A D let a circle be described (41). The part A E of the greater cut off by this circle will be equal to the less C.

### Demonstration.

For A E and A D are equal, being radii of the same circle (17); and C and A D are equal by the construction. Hence A E and C are equal.

By a similar construction, the less might be produced until it equal the greater. From an extremity of the less let a line equal to the greater be drawn, and a circle be described with this line as radius. Let the less be produced to meet this circle.

## Euclid, Elements of Geometry, Book I, Proposition 4

### Proposition 4[Heath's Edition]

If two triangles have the two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal angles subtend.

Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF, respectively, namely AB to DE and AC to DF, and the angle BAC equal to the angle EDF.

I say that the base BC is also equal to the base EF, the triangle ABC will be equal to the triangle DEF, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, that is, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

For, if the triangle ABC be applied to the triangle DEF,
and if the point A be placed on the point D
and the straight line AB on DE,
then the point B will also coincide with E, because AB is equal to DE.

Again, AB coinciding with DE, the straight line AC will also coincide with DF, because the angle BAC is equal to the angle EDF;
hence the point C will also coincide with the point F, because AC is again equal to DF.

But B also coincided with E;
hence the base BC will coincide with the base EF.

[For if, when B coincides with E and C with F, the base BC does not coincide with the base EF, two straight lines will enclose a space: which is impossible.

Therefore the base BC will coincide with EF] and will be equal to it. [C.N. 4]

Thus the whole triangle ABC will coincide with the whole triangle DEF,
and will be equal to it.

And the remaining angles will also coincide with the remaining angles and will be equal to them,
the angle ABC to the angle DEF,
and the angle ACB to the angle DFE.

Therefore etc.
(Being) what it was required to prove.

Proposition IV. Theorem. [Lardner's Edition]

 (62) If two triangles (B A C and E D F) have two sides (B A and A C) in the one respectively equal to two sides (E D and D F) in the other, and the angles (A and D) included by those sides also equal; the bases or remaining sides (B C and E F) will be equal, also the angles (B and C) at the base of the one will be respectively equal to those (E and F) at the base of the other which are opposed to the equal sides (i. e. B to E and C to F).

Let the two triangles be conceived to be so placed that the vertex of one of the equal angles D shall fall upon that of the other A, that one of the sides D E containing the given equal angles shall fall upon the side A B in the other triangle to which it is equal, and that the remaining pair of equal sides A C and D F shall lie at the same side of those A B and D E which coincide.

Since then the vertices A and D coincide, and also the equal sides A B and D E, the points B and E must coincide. (If they did not the sides A B and D E would not be equal.) Also, since the side D E falls on A B, and the sides A C and D F are at the same side of A B, and the angles A and D are equal, the side D F must fall upon A C; (for otherwise the angles A and D would not be equal.)

Since the side D F falls on A C, and they are equal, the extremity F must fall on C. Since the extremities of the bases B C and E F coincide, these lines themselves must coincide, for if they did not they would include a space (52). Hence the sides B C and E F are equal (50).

Also, since the sides E D and E F coincide respectively with B A and B C, the angles E and B are equal (50), and for a similar reason the angles F and C are equal.

Since the three sides of the one triangle coincide respectively with the three sides of the other, the triangles themselves coincide, and are therefore equal (50).

In the demonstration of this proposition, the converse of the eighth axiom (50) is assumed. The axiom states, that ‘if two magnitudes coincide they must be equal.’ In the proposition it is assumed, that if they be equal they must under certain circumstances coincide. For when the point D is placed on A, and the side D E on A B, it is assumed that the point E must fall on B, because A B and D E are equal. This may, however, be proved by the combination of the eighth and ninth axioms; for if the point E did not fall upon B, but fell either above or below it, we should have either E D equal to a part of B A, or B A equal to a part of E D. In either case the ninth axiom would be contradicted, as we should have the whole equal to its part.

The same principle may be applied in proving that the side D F will fall upon A C, which is assumed in Euclid's proof.

In the superposition of the triangles in this proposition, three things are to be attended to:

1° The vertices of the equal angles are to be placed one on the other.

2° Two equal sides to be placed one on the other.

3° The other two equal sides are to be placed on the same side of those which are laid one upon the other.

From this arrangement the coincidence of the triangles is inferred.

It should be observed, that this superposition is not assumed to be actually effected, for that would require other postulates besides the three already stated; but it is sufficient for the validity of the reasoning, if it be conceived to be possible that the triangles might be so placed. But the same principle of superposition, the following theorem must be easily demonstrated, ‘If two triangles have two angles in one respectively equal to two angles in the other, and the sides lying between those angles also equal, the remaining sides and angles will be equal, and also the triangles themselves will be equal.’ See prop. xxvi.

This being the first theorem in the Elements, it is necessarily deduced exclusively from the axioms, as the first problem must be from the postulates. Subsequent theorems and problems will be deduced from those previously established.

[Proposition IV. Theorem]. [Playfair's Edition]

### PROP. IV. and VIII. B. I.

The fourth and eighth propositions of the first book are the foundation of all that follows with respect to the comparison of triangles. They are demonstrated by what is called the method of supraposition, that is, by laying the one triangle upon the other, and proving that they must coincide. To this some objections have been made, as if it were ungeometrical to suppose one figure to be removed from its place and applied to another figure. “The laying,” says Mr Thomas Simpson in his Elements, “of one figure upon another, whatever evidence it may afford, is a mechanical consideration, “and depends on no postulate.” It is not clear what Mr Simpson meant here by the word mechanical; but he probably intended only to say, that the method of supraposition involves the idea of motion, which belongs rather to mechanics than geometry; for I think it is impossible that such a Geometer as he was could mean to assert, that the evidence derived from this method is like that which arises from the use of instruments, and of the same kind with what is furnished by experience and observation. The demonstrations of the fourth and eighth, as they are given in Euclid, are as certainly a process of pure reasoning, depending solely on the idea of equality, as established in the 8th Axiom, as any thing in geometry. But, if still the removal of the triangle from its place be considered as creating a difficulty, and as inelegant, because it involves an idea, that of motion, not essential to geometry, this defect may be entirely remedied, provided that, to Euclid's three postulates, we be allowed to add the following, viz. This if there be two equal straight lines, and if any figure whatsoever be constituted on the one, a figure every way equal to it may be constituted on the other. Thus, if AB and DE be two equal straight lines, and ABC a triangle on the base AB, a triangle DEF every way equal to ABC may be supposed to be constituted on DE as a base. By this it is not meant to assert that the method of describing the triangle DEF is actually known, but merely that the triangle DEF may be conceived to exist in all respects equal to the triangle ABC. Now, there is no truth whatsoever that is better entitled than this to be ranked among the Postulates or Axioms of geometry; for the straight lines AB and DE being every way equal, there can be nothing belonging to the one that may not also belong to the other.

On the strength of this postulate the fourth Proposition is thus demonstrated.

If ABC, DEF be two triangles, such that the two sides AB and AC of the one are equal to the two ED and EF of the other, and the angle BAC, contained by the sides AB, AC of the one, equal to the angle EDF, contained by the sides ED, DF of the other; the triangles If ABC, EDF are every way equal.

On AB let a triangle be constituted every way equal to the triangle ABC, it is is evident that the proposition is true, for it is equal to DEF by hypothesis, and to ABC, because it coincids with it; wherefore ABC, DEF are equal to one another. But if it does not coincide with ABC, let it have the position ABG; and first suppose G not to fall on AC; then the angle BAG is not equal to the angle BAC. But the angle BAG is equal to the angle EDF therefore EDF and ABC are not equal, and they are also equal by hypothesis, which is impossible. Therefore the point G must fall upon AC; now, if it fall upon AC but not at C, then AG is not equal to AC; but AG is equal to D ; therefore DF and AC are not equal, and they are also equal by supposition, which is impossible. Therefore G must coincide with C, and the triangle AGB with the triangle ACB. But But AGB is every way equal to But DEF, therefore But ACB and But DEF are also every way equal. Q. E. D.

[Note continues.]

## Euclid, Elements of Geometry, Book I, Proposition 5

### Proposition 5[Heath's Edition]

In isosceles triangles the angles at the base are equal to one another, and, if the equal straight lines be produced further; the angles under the base will be equal to one another.

Let ABC be an isosceles triangle having the side AB equal to the side AC;
and let the straight lines BD, CE, be produced further in a straight line with AB, AC. [Post. 2]

I say that the angle ABC is equal to the angle ACB, and the angle CBD to the angle BCE.

Let a point F be taken at random on BD;
from AE the greater let AG be cut off equal to AF the less; [I. 3]
and let the straight lines FC, GB be joined. [Post. 1]

Then, since AF is equal to AG and AB to AC,
the two sides FA, AC are equal to the two sides GA, AB respectively;
and they contain a common angle, the angle FAG.

Therefore the base FC is equal to the base GB,
and the triangle AFC is equal to the triangle AGB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend,
that is, the angle ACF to the angle ABG,
and the angle AFC to the angle AGB. [I. 4]

And, since [I. 4]

Again, let sides subtending equal angles be equal, as AB to DE;
I say again that the remaining sides will be equal to the remaining sides, namely AC to DF and BC to EF, and further the remaining angle BAC is equal to the remaining angle EDF.

For if BC is unequal to EF, one of them is greater.

Let BC be greater, if possible, and let BH be made equal to EF; let AH be joined.

Then, since BH is equal to EF, and AB to DE,
the two sides AB, BH are equal to the two sides DE, EF respectively, and they contain equal angles;
therefore the base AH is equal to the base DF,
and the triangle ABH is equal to the triangle DEF,
and the remaining angles will be equal to the remaining angles, namely those which equal sides subtend; [I. 4]
therefore the angle BHA is equal to the angle EFD.

But the angle EFD is equal to the angle BCA;
therefore, in the triangle AHC, the exterior angle BHA is equal to the interior and opposite angle BCA:
which is impossible. [I. 16]
Therefore BC is not unequal to EF,
and is therefore equal to it.

But AB is also equal to DE;
therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and they contain equal angles;
therefore the base AC is equal to the base DF,
the triangle ABC equal to the triangle DEF,
and the remaining angle BAC equal to the remaining angle EDF. [I. 4]

Therefore etc. Q.E.D.

Proposition V. Theorem. [Lardner's Edition]

 (63) The angles (B, C) opposed to the equal sides (A C and A B) of an isosceles triangle are equal, and if the equal sides be produced through the extremities (B and C) of the third side, the angles (D B C and E C B) formed by their produced parts and the third side are equal.

Let the equal sides A B and A C be produced through the extremities B, C, of the third side, and in the produced part B D of either let any point F be assumed, and from the other let A G be cut off equal to A F (III). Let the points F and G so taken on the produced sides be connected by right lines F C and B G with the alternate extremities of the third side of the triangle.

In the triangles F A C and G A B the sides F A and A C are respectively equal to G A and A B, and the included angle A is common to both triangles. Hence (IV), the line F C is equal to B G, the angle A F C to the angle A G B, and the angle A C F to the angle A B G. If from the equal lines A F and A G, the equal sides A B and A C be taken, the remainders B F and C G will be equal. Hence, in the triangles B F C and C G B, the sides B F and F C are respectively equal to C G and G B, and the angles F and G included by those sides are also equal. Hence (IV), the angles F B C and G C B, which are those included by the third side B C and the productions of the equal sides A B and A C, are equal. Also, the angles F C B and G B C are equal. If these equals be taken from the angles F C A and G B A, before proved equal, the remainders, which are the angles A B C and A C B opposed to the equal sides, will be equal.

 (64) Cor.—Hence, in an equilateral triangle the three angles are equal; for by this proposition the angles opposed to every two equal sides are equal.

## Euclid, Elements of Geometry, Book I, Proposition 6

### Proposition 6[Heath's Edition]

If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.

Let ABC be a triangle having the angle ABC equal to the angle ACB;
I say that the side AB is also equal to the side AC.

For, if AB is unequal to AC, one of them is greater.

Let AB be greater; and from AB the greater let DB be cut off equal to AC the less;
let DC be joined.

Then, since DB is equal to AC, and BC is common,
the two sides DB, BC are equal to the two sides AC, CB respectively;
and the angle DBC is equal to the angle ACB;
therefore the base DC is equal to the base AB,
and the triangle DBC will be equal to the triangle ACB,
the less to the greater;
which is absurd.

Therefore AB is not unequal to AC;
it is therefore equal to it.

Therefore etc. Q.E.D.

Proposition VI. Theorem. [Lardner's Edition]

 (65) If two angles (B and C) of a triangle (B A C) be equal, the sides (A C and A B) opposed to them are also equal.

For if the sides be not equal, let one of them A B be greater than the other, and from it cut off D B equal to A C (III), and draw C D.

Then in the triangles D B C and A C B, the sides D B and B C are equal to the sides A C and C B respectively, and the angles D B C and A C B are also equal; therefore (IV) the triangles themselves D B C and A C B are equal, a part equal to the whole, which is absurd; therefore neither of the sides A B or A C is greater than the other; there are therefore equal to one another.

 (66) Cor.—Hence every equiangular triangle is also equilateral, for the sides opposed to every two equal angles are equal.

In the construction for this proposition it is necessary that the part of the greater side which is cut off equal to the less, should be measured upon the greater side B A from vertex (B) of the equal angle, for otherwise the fourth proposition could not be applied to prove the equality of the part with the whole.

It may be observed generally, then when a part of one line is cut off equal to another, it should be distinctly specified from which extremity the part is to be cut.

This proposition is what is called by logicians the converse of the fifth. It cannot however be inferred from it by the logical operation called conversion; because, by the established principles of Aristotelian logic, an universal affirmative admits no simple converse. This observation applies generally to those propositions in the Elements which are converses of preceding ones.

The demonstration of the sixth is the first instance of indirect proof which occurs in the Elements. The force of this species of demonstration consists in showing that a principle is true, because some manifest absurdity would follow from supposing it to be false.

This kind of proof is considered inferior to direct demonstration, because it only proves that a thing must be so, but fails in showing why it must be so; whereas direct proof not only shows that the thing is so, but why it is so. Consequently, indirect demonstration is never used, except where no direct proof can be had. It is used generally in proving principles which are nearly self-evident, and in the Elements if oftenest used in establishing the converse propositions. Examples will be seen in the 14th, 19th, 25th and 40th propositions of this book.

## Euclid, Elements of Geometry, Book I, Proposition 7

### Proposition 7[Heath's Edition]

Given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there cannot be constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.

For, if possible, given two straight lines AC, CB constructed on the straight line AB and meeting at the point C, let two other straight lines AD, DB be constructed on the same straight line AB, on the same side of it, meeting in another point D and equal to the former two respectively, namely each to that which has the same extremity with it, so that CA is equal to DA which has the same extremity A with it, and CB to DB which has the same extremity B with it; and let CD be joined.

Then, since AC is equal to AD,
the angle ACD is also equal to the angle ADC; [I. 5]
therefore the angle ADC is greater than the angle DCB;
therefore the angle CDB is much greater than the angle DCB.

Again, since CD is equal to DB,
the angle CDB is also equal to the angle DCB.
But it was also proved much greater than it:
which is impossible.

Therefore etc. Q.E.D.

Proposition VII. Theorem. [Lardner's Edition]

 (67) On the same right line (A B), and on the same side of it, there cannot be constructed two triangles, (A C B, A D B) whose conterminous sides (A C and A D, B C and B D) are equal.

If it be possible, let the two triangles be constructed, and,

First,—Let the vertex of each of the triangles be without the other triangle, and draw C D.

Because the sides A D and A C of the triangle C A D are equal (hyp., [see note]), the angles A C D and A D C are equal (V); but A C D is greater than B C D (51), therefore A D C is greater than B C D; but the angle B D C is greater than A D C (51), and therefore B D C is greater than B C D; but in the triangle C B D, the sides B C and B D are equal (hyp.), therefore the angles B D C and B C D are equal (V); but the angle B D C has been proved to be greater than B C D, which is absurd: therefore the triangles constructed upon the same right line cannot have their conterminous sides equal, when the vertex of each of the triangles is without the other.

Secondly,—Let the vertex D of one triangle be within the other; produce the sides A C and A D, and join C D.

Because the sides A C and A D of the triangle C A D are equal (hyp.), the angles E C D and F D C are equal (V); but the angle B D C is greater than F D C (51), therefore greater than E C D; but E C D is greater than B C D (51), and therefore B D C is greater than B C D; but in the triangle C B D, the sides B C and B D are equal (hyp.), therefore the angles B D C and B C D are equal (V); but the angle B D C has been proved to be greater than B C D, which is absurd: therefore triangles constructed on the same right line cannot have their conterminous sides equal, if the vertex of one of them is within the other.

Thirdly,—Let the vertex D of one triangle be on the side A B of the other, and it is evident that the sides A B and B D are not equal.

Therefore in no case can two triangles, whose conterminous sides are equal, be constructed at the same side of the given line.

This proposition seems to have been introduced into the Elements merely for the purpose of establishing that which follows it. The demonstration is that form of argument which logicians call a dilemma, and a species of argument which seldom occurs in the Elements. If two triangles whose conterminous sides are equal could stand on the same side of the same base, the vertex of the one must necessarily either fall within the other or without it, or on one of the sides of it: accordingly, it is successively proved in the demonstration, that to suppose it in any of these positions would lead to a contradiction in terms. It is not supposed that the vertex of the one could fall on the vertex of the other; for that would be supposing the two triangles to be one and the same, whereas they are, by hypothesis, different.

In the Greek text there is but one (the first) of the cases of this proposition given. It is however conjectured, that the second case must have been formerly in the text, because it is the only instance in which Euclid uses that part of the fifth proposition which proves the equality of the angles below the base. It is argued, that there must have been some reason for introducing into the fifth a principle which follows at once from the thirteenth; and that none can be assigned except the necessity of the principle in the second case of the seventh. The third case required to be mentioned only to preserve the complete logical form of the argument.

## Euclid, Elements of Geometry, Book I, Proposition 8

### Proposition 8[Heath's Edition]

If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they will also have the angles equal which are contained by the equal straight lines.

Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE and AC to DF; and let them have the base BC equal to the base EF;
I say that the angle BAC is also equal to the angle EDF.

For if the triangle ABC be applied to the triangle DEF, and if the point B be placed on the point E and the straight line BC on EF,
the point C will also coincide with F,
because BC is equal to EF.

Then, BC coinciding with EF,
BA, AC will also coincide with ED, DF;
for, if the base BC coincides with the base EF, and the sides BA, AC do not coincide with ED, DF but fall beside them as EG, GF,
then, given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there will have been constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it.

But they cannot be so constructed. [I. 7]

Therefore it is not possible that, if the base BC be applied to the base EF, the sides BA, AC should not coincide with ED, DF;
they will therefore coincide,
so that the angle BAC will also coincide with the angle EDF, and will be equal to it.

If therefore etc. Q.E.D.

Proposition VIII. Theorem. [Lardner's Edition]

 (68) If two triangles (A B C and E F D) have two sides of the one respectively equal to two sides of the other (A B to E F and C B to D F), and also have the base (A C) equal to the base (E D), then the angles (B and F) contained by the equal sides are equal.

For if the equal bases A C, E D be conceived to be placed one upon the other, so that the triangles shall lie at the same side of them, and that the equal sides A B and E F, C B and D F be conterminous, the vertex B must fall on the vertex F; for to suppose them not coincident would contradict the seventh proposition. The sides B A and B C being therefore coincident with F E and F D, the angles B and F are equal.

 (69) It is evident that in this case all the angles and sides of the triangles are respectively equal each to each, and that the triangles themselves are equal. This appears immediately by the eighth axiom.

In order to remove from the threshold of the Elements a proposition so useless, and, to the younger students, so embarrassing as the seventh, it would be desirable that the eighth should be established independently of it. There are several ways in which this might be effected. The following proof seems liable to no objection, and establishes the eighth by the fifth.

Let the two equal bases be so applied one upon the other that the equal sides shall be conterminous, and that the triangles shall lie at opposite sides of them, and let a right line be conceived to be drawn joining the vertices.

1° Let this line intersect the base.

Let the vertex F fall at G, the side E F in the position A G, and D F in the position C G. Hence B A and A G being equal, the angles G B A and B G A are equal (V). Also C B and C G being equal, the angles C G B and C B G are equal (V). Adding these equals to the former, the angles A B C and A G C are equal; that is, the angles E F D and A B C are equal.

2° Let the line G B fall outside the coincident bases.

The angles G B A and B G A, and also B G C and G B C are proved equal as before; and taking the latter from the former, the remainders, which are the angles A G C and A B C, are equal, but A G C is the angle F.

3° Let the line B G pass through either extremity of the base.

In this case it follows immediately (V) that the angles A B C and A G C are equal; for the lines B C and C G amust coincide with B G, since each has two points upon it (52).

Hence in every case the angles B and F are equal.

This proposition is also sometimes demonstrated as follows.

Conceive the triangle E F D to be applied to A B C, as in Euclid's proof. Then because E F is equal to A B, the point F must be in the circumference of a circle described with A as centre, and A B as radius. And for the same reason, F must be on a circumference with the centre C, and the radius C B. The vertex must therefore be at the point where these circles meet. But the vertex B must be also at that point; wherefore &c.

## Euclid, Elements of Geometry, Book I, Proposition 9

### Proposition 9[Heath's Edition]

To bisect a given rectilineal angle.

Let the angle BAC be the given rectilineal angle.

Thus it is required to bisect it.

Let a point D be taken at random on AB;
let AE be cut off from AC equal to AD; [I. 3]
let DE be joined, and on DE let the equilateral triangle DEF be constructed;
let AF be joined.

I say that the angle BAC has been bisected by the straight line AF.

For, since AD is equal to AE,
and AF is common,
the two sides DA, AF are equal to the two sides EA, AF respectively.
And the base DF is equal to the base EF;
therefore the angle DAF is equal to the angle. EAF [I. 8]

Therefore the given rectilineal angle BAC has been bisected by the straight line AF. Q.E.F.

Proposition IX. Problem. [Lardner's Edition]

 (70) To bisect a given rectilinear angle (B A C).

### Solution.

Take any point D in the side A B, and from A C cut off A E equal to A D (III), draw D E, and upon it describe an equilateral triangle D F E (I) at the side remote from A. The right line joining the points A and F bisects the given angle B A C.

### Demonstration.

Because the sides A D and A E are equal (const.), and the side A F is common to the triangles F A D F A E, and the base F D is also equal to F E (const.); the angles D A F and E A F are equal (VIII), and therefore the right line bisects the given angle.

By this proposition an angle may be divided into 4. 8, 16 &c. equal parts, or, in general, into any number of equal parts which is expressed by a power of two.

It is necessary that the equilateral triangle be constructed on a different side of the joining line D E from that on whihc the given angle is placed, lest the vertex F of the equilateral triangle should happen to coincide with the vertex A of the given angle; in which case there would be no joining line F A, and therefore no solution.

In these cases, however, in which the vertex of the equilateral triangle does not coincide with that of the given angle, the problem can be solved by constructing the equilateral triangle on the same side of the joining line D E with the given angle. Separate demonstrations are necessary for the two positions which the vertices may assume.

1. Let the vertex of the equilateral triangle fall within that of the given angle.

The demonstration already given will apply to this without any modification.

2. Let the vertex of the given angle fall within the equilateral triangle.

The line F A produced will in this case bisect the angle; for the three sides of the triangle D F A are respectively equal to those of the triangle E F A. Hence the angles D F A and E F A are equal (VIII). Also, in the triangles D F G and E F G the sides D F and E F are equal, the side G F is common, and the angles D F G and E F G are equal; hence (IV) the bases D G and E G are equal, A G is common, and the angles at G are equal; hence (IV) the angles D A G and E A G are equal, and therefore the angle B A C is bisected by A G.

It is evident, that an isosceles triangle constructed on the joining line D E would equally answer the purpose of the solution.

## Euclid, Elements of Geometry, Book I, Proposition 10

### Proposition 10[Heath's Edition]

To bisect a given finite straight line.

Let AB be the given finite straight line.

Thus it is required to bisect the given straight line AB.

Let the equilateral triangle ABC be constructed on it, [I. 1]
and let the angle ACB be bisected by the straight line CD; [I. 9]
I say that the straight line AB has been bisected at the point D.

For, since AC is equal to CB, and CD is common,
the two sides AC, CD are equal to the two sides BC, CD respectively;
and the angle ACD is equal to the angle BCD;
therefore the base AD is equal to the base BD; [I. 4]

Therefore the given finite straight line AB has been bisected at D. Q.E.F.

Proposition X. Problem. [Lardner's Edition]

 (71) To bisect a given right line (A B).

### Solution.

Upon the given line A B describe an equilateral triangle A C B (I), bisect the angle A C B by the right line C D (IX); this line bisects the given line in the point D.

### Demonstration.

Because the sides A C and C B are equal (const.), and C D common to the triangles A C D and B C D, and the angles A C D and B C D also equal (const.); therefore (IV) the bases A D D B are equal, and the right line A B is bisected in the point D.

In this and the following proposition an isosceles triangle would answer the purposes of the solution equally with an equilateral. In fact, in the demonstrations the triangle is contemplated merely as isosceles: for nothing is inferred from the equality of the base with the sides.

## Euclid, Elements of Geometry, Book I, Proposition 11

### Proposition 11[Heath's Edition]

To draw a straight line at right angles to a given straight line from a given point on it.

Let AB be the given straight line, and C the given point on it.

Thus it is required to draw from the point C a straight line at right angles to the straight line AB.

Let a point D be taken at random on AC;
let CE be made equal to CD; [I. 3]
on DE let the equilateral triangle FDE be constructed,
and let FC be joined;
I say that the straight line FC has been drawn at right angles to the given straight line AB from C the given point on it.

For, since DC is equal to CE,
and CF is common,
the two sides DC, CF are equal to the two sides EC, CF respectively;
and the base DF is equal to the base FE;
therefore the angle DCF is equal to the angle ECF; [I. 8]

But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [Def. 10]
therefore each of the angles DCF, FCE is right.

Therefore the straight line CF has been drawn at right angles to the given straight line AB from the given point C on it. Q.E.F.

Proposition XI. Problem. [Lardner's Edition]

 (72) From a given point (C) in a given right line (A B) to draw a perpendicular to the given line.

### Solution.

In the given line take any point D and make C E equal to C D (III); upon D E describe an equilateral triangle D F E (I); draw F C, and it is perpendicular to the given line.

### Demonstration.

Because the sides D F and D C are equal to the sides E F and E C (const.), and C F is common to the triangles D F C and E F C, therefore (VIII) the angles opposite to the equal sides D F and E F are equal, and therefore F C is perpendicular to the given right line A B at the point C.

 Cor.—By help of this problem it may be demonstrated, that two straight lines cannot have a common segment.

It it be possible, let the two straight lines A B C, A B D have the segment A B common to both of them. From the point B draw B E at right angles to A B; and because A B C is a straight line, the angle C B E is equal to the angle E B A; in the same manner, because A B D is a straight line, the angle D B E is equal to the angle E B A; wherefore the angle D B E is equal to the angle C B E, the less to the greater, which is impossible; therefore the two straight lines cannot have a common segment.

If the given point be at the extremity of the given right line, it must be produced, in order to draw the perpendicular by this construction.

In a succeeding article, the student will find a method of drawing a perpendicular through the extremity of a line without producing it.

The corollary to this proposition is useless, and is omitted in some editions.

It is equivalent to proving that a right line cannot be produced through its extremity in more than one direction, or that it has but one production.

## Euclid, Elements of Geometry, Book I, Proposition 12

### Proposition 12[Heath's Edition]

To a given infinite straight line, from a given point which is not on it, to draw a perpendicular straight line.

Let AB be the given infinite straight line, and C the given point which is not on it;
thus it is required to draw to the given infinite straight line AB, from the given point C which is not on it, a perpendicular straight line.

For let a point D be taken at random on the other side of the straight line AB, and with centre C and distance CD let the circle EFG be described; [Post. 3]
let the straight line EG be bisected at H,
and let the straight lines CG, CH, CE be joined. [Post. 1]
I say that CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it.

For, since GH is equal to HE,
and HC is common,
the two sides GH, HC are equal to the two sides EH, HC respectively;
and the base CG is equal to the base CE;
therefore the angle CHG is equal to the angle EHC. [I. 8]

But when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands. [Def. 10]

Therefore CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it. Q.E.F.

Proposition XII. Problem. [Lardner's Edition]

 (73) To draw a perpendicular to a given indefinite right line (A B). from a point (C) given without it.

### Solution.

Take any point X on the other side of the given line, and from the centre C with the radius C X describe a circle cutting the given line in E and F. Bisect E F in D (X), and draw from the given point to the point of bisection the right line C D; this line is the required perpendicular.

### Demonstration.

For draw C E and C F, and in the triangles E D C and F D C the sides E C and F C, and E D and F D, are equal (const.) and C D common; therefore (VIII) the angles E D C and F D C opposite to the equal sides E C and F C are equal, and therefore D C is perpendicular to the line A B (11).

In this proposition it is necessary that the right line A B be indefinite in length, for otherwise it might happen that the circle described with the centre C and the radius C X might not intersect it in two points, which is essential to the solution of the problem.

It is assumed in the solution of this problem, that the circle will intersect the right line in two points. The centre of the circle being on one side of the given right line, and a part of the circumference (X) on the other, it is not difficult to perceive that a part of the circumference must also be also on the same side of the given line with the centre, and since the circle is a continued line it must cross the right line twice. The properties of the circle form the subject of the third book, and those which are assumed here will be established in that part of the Elements.

The following questions will afford the student useful exercise in the application of the geometrical principles which have been established in the last twelve propositions.

 (74) In an isosceles triangle the right line which bisects the vertical angle also bisects the base, and is perpendicular to the base.

For the two triangles into which it divides the isosceles, there are two sides (those of the isosceles) equal, and a side (the bisector) common, and the angles included by those sides equal, being the parts of the bisected angle; hence (IV) the remaining sides and angles are respectively equal; that is, the parts into which the base is divided by the bisector are equal, and the angle which the bisector makes with the base are equal. Therefore it bisects the base, and is perpendicular to it.

It is clear that the isosceles triangle itself is bisected by the bisector of its vertical angle, since the two triangles are equal.

 (75) It follows also, that in an isosceles triangle the line which is drawn from the vertex to the middle point of the base bisects the vertical angle, and is perpendicular to the line.

For in this case the triangle is divided into two triangles, which have their three sides respectively equal each to each, and the property is established by (VIII)

 (76) If in a triangle the perpendicular from the vertex on the base bisect the base, the triangle is isosceles.

For in this case in the two triangles into which the whole is divided by the perpendicular, there are two sides (the parts of the base) equal, one side (the perpendicular) common, and the included angles equal, being right. Hence (IV) the sides of the triangle are equal.

 (77) To find a point which is equidistance from the three vertical points of a triangle A B C.

Bisect the sides A B and B C at D and E (X), through the points D and E draw perpendiculars, and produce them until they meet at F. The point F is at equal distances from A, B and C.

For draw F A, F B, F C. B F A is isosceles by (76), and for the same reason B F C is isosceles. Hence it is evident that F A, F C, and F B are equal.

 (78) Cor.—Hence F is the centre, and F A the radius of a circle circumscribed about the triangle.
 (79) In a quadrilateral formed by two isosceles triangles A C B and A D B constructed on different sides of the same base, the diagonals intersect at right angles, and that which is the common base of the isosceles triangles is bisected by the other.

For in the triangles C A D and C B D the three sides are equal each to each, and therefore (VIII) the angles A C E and B C E are equal. The truth of the proposition therefore follows from (74)

 (80) Hence it follows that the diagonals of a lozenge bisect each other at right angles.
 (81) It follows from (76) that if the diagonals of a quadrilateral bisect each other at right angles it is a lozenge.

## Euclid, Elements of Geometry, Book I, Proposition 13

### Proposition 13[Heath's Edition]

If a straight line set up on a straight line make angles, it will make either two right angles or angles equal to two right angles.

For let any straight line AB set up on the straight line CD make the angles CBA, ABD;
I say that the angles CBA, ABD are either two right angles or equal to two right angles.

Now, if the angle CBA is equal to the angle ABD, they are two right angles. [Def. 10]

But, if not, let BE be drawn from the point B at right angles to CD; [I. 11]
therefore the angles CBE, EBD are two right angles.

Then, since the angle CBE is equal to the two angles CBA, ABE,
let the angle EBD be added to each;
therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. [C.N. 2]

Again, since the angle DBA is equal to the two angles DBE, EBA,
let the angle ABC be added to each;
therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC. [C.N. 2]

But the angles CBE, EBD were also proved equal to the same three angles;
and things which are equal to the same thing are also equal to one another; [C.N. 1]
therefore the angles CBE, EBD are also equal to the angles DBA, ABC.
But the angles CBE, EBD are two right angles;
therefore the angles DBA, ABC are also equal to two right angles.

Therefore etc. Q.E.D.

Proposition XIII. Theorem. [Lardner's Edition]

 (82) When a right line (A B) standing upon another (D C) makes angles with it, they are either two right angles, or together equal to two right angles.

If the right line A B is perpendicular to D C, the angles A B C and A B D are right (11). If not, draw B E perpendicular to D C (XI), and it is evident that the angles C B A and A B D together are equal to the angles C B E and E B D, and therefore to two right angles.

The words ‘makes angles with it,’ are introduced to exclude the case in which the line A B is at the extremity of B C.

(83)   From this proposition it appears, that if several right lines stand on the same right line at the same point, and make angles with it, all the angles taken together are equal to two right angles.

Also if two right lines intersecting one another make angles, these angles taken together are equal to four right angles.

The lines which bisect the adjacent angles A B C and A B D are at right angles; for the angle under these lines is evidently half the sum of the angles A B C and A B D.

If several right lines diverge from the same point, the angles into which they divide the surrounding space are together equal to four right angles.

(84)   When two angles as A B C and A B D are togther equal to two right angles, they are said to be supplemental, and one is called the supplement of the other.

(85)   If two angles as C B A and E B A are together equal to a right angle, they are said to be complemental, one one is said to be the complement of the other.

## Euclid, Elements of Geometry, Book I, Proposition 14

### Proposition 14[Heath's Edition]

If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with one another.

For with any straight line AB, and at the point B on it, let the two straight lines BC, BD not lying on the same side make the adjacent angles ABC, ABD equal to two right angles;
I say that BD is in a straight line with CB.

For, if BD is not in a straight line with BC let BE be in a straight line with CB.

Then, since the straight line AB stands on the straight line CBE,
the angles ABC, ABE are equal to two right angles; [I. 13]
But the angles ABC, ABD are also equal to two right angles;
therefore the angles CBA, ABE are equal to the angles CBA, ABD. [Post. 4 and C.N. 1]

Let the angle CBA be subtracted from each;
therefore the remaining angle ABE is equal to the remaining angle ABD,
the less to the greater: which is impossible.
Therefore BE is not in a straight line with CB.

Similarly we can prove that neither is any other straight line except BD.
Therefore CB is in a straight line with BD.

Therefore etc. Q.E.D.

Proposition XIV. Theorem. [Lardner's Edition]

 (86) If two right lines (C B and B D) meeting another right line (A B) at the same point (B), and at opposite sides, make angles with it which are together equal to two right angles, those right angles (C B and B D) form one continued right line.

For if possible, let B E and not B D be the continuation of the right line C B, then the angles C B A and C B E are are equal to two right angles (XIII), but C B A and A B D are also equal to two right angles, by hypothesis, therefore C B A and A B D taken together are equal to C B A and A B E; take away from these equal quantities C B A which is common to both, and A B E shall be equal to A B D, a part to the whole, which is absurd; therefore B E is not the continuation of C B, and in the same manner it can be proved, that no other line except B D is the continuation of it, therefore B D forms with B C one continued right line.

In the enunciation of this proposition, the student should be cautious not to overlook the condition that the two right lines C B and B E forming angles, which are together equal to two right angles, with B A lie at opposite sides of B A. They might form angles together equal to two right angles with B A, yet not lie in the same continued line, if as in this figure they lay at the same side of it. It is assumed in this proposition that the line X Y has a production. This is however granted by Postulate~2.

## Euclid, Elements of Geometry, Book I, Proposition 15

### Proposition 15[Heath's Edition]

If two straight lines cut one another, they make the vertical angles equal to one another.

For let the straight lines AB, CD cut one another at the point E;
I say that the angle AEC is equal to the angle DEB,
and the angle CEB to the angle AED.

For, since the straight line AE stands on the straight line CD, making the angles CEA, AED,
the angles CEA, AED are equal to two right angles. [I. 13]

Again, since the straight line DE stands on the straight line AB, making the angles AED, DEB,
the angles AED, DEB are equal to two right angles [I. 13]

But the angles CEA, AED were also proved equal to two right angles;
therefore the angles CEA, AED are equal to the angles AED, DEB. [Post. 4 and C.N. 1]
Let the angle AED be subtracted from each;
therefore the remaining angle CEA is equal to the remaining angle BED. [C.N. 3]

Similarly it can be proved that the angles CEB, DEA are also equal.

Therefore, etc. Q.E.D.

[PORISM. From this it is manifest that, if two straight lines cut one another, they will make the angles at the point of section equal to four right angles.]

Proposition XV. Theorem. [Lardner's Edition]

 (87) If two right lines (A B and C D) intersect one another, the vertical angles are equal (C E A to B E D, and C E B to A E D).

Because the right line C E stands upon the right line A B, the angle A E C together with the angle C E B is equal to two right angles (XIII); and because the right line B E stands on the right line C D, the angle C E B together with the angle B E D is equal to two right angles (XIII); therefore A E C and C E B together are equal to C E B and B E D; take away the common angle C E B, and the remaining angle A E C is equal to B E D.

This proof may shortly be expressed by saying, that opposite angles are equal, because they have a common supplement (84).

It is evident that angles which have a common supplement or complement (85) are equal, and that if they be equal, their supplements and complements must also be equal.

(88)   The converse of this proposition may easily be proved, scil. If four lines meet at a point, and the angles vertically opposite be equal, each alternate pair of lines will be in the same right line. For if C E A be equal to B E D, and also C E B to A E D, it follows that C E A and C E B together are equal to B E D and A E D together. But all the four are together euqal to four right angles (83), and therefore C E A and C E B are together equal to two right angles, therefore (XIV) A E and A B are in one continued line. In like manner it may be proved, that C E and D E are in one line.

## Euclid, Elements of Geometry, Book I, Proposition 16

### Proposition 16[Heath's Edition]

In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles.

Let ABC be a triangle, and let one side of it BC be produced to D;
I say that the exterior angle ACD is greater than either of the interior and opposite angles CBA, BAC.

Let AC be bisected at E [I. 10] , and let BE be joined and produced in a straight line to F;
let EF be made equal to BE [I. 3] , let FC be joined [Post. 1] , and let AC be drawn through to G. [Post. 2]

Then, since AE is equal to EC, and BE to EF,
the two sides AE, EB are equal to the two sides CE, EF respectively;
and the angle AEB is equal to the angle FEC,
for they are vertical angles. [I. 15]
Therefore the base AB is equal to the base FC,
and the triangle ABE is equal to the triangle CFE,
and the remaining angles are equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4]
therefore the angle BAE is equal to the angle ECF.

But the angle ECD is greater than the angle ECF; [C.N. 5]
therefore the angle ACD is greater than the angle BAE.

Similarly also, if BC be bisected, the angle BCG, that is, the angle ACD [I. 15] , can be proved greater than the angle ABC as well.

Therefore etc. Q.E.D.

Proposition XVI. Theorem. [Lardner's Edition]

 (89) If one side (B C) of a triangle (B A C) be produced, the external angle (A C D) is greater than either of the internal opposite angles (A or B.)

For bisect the side A C in E (X), draw B E and produce it until E F be equal to B E (III), and join F C.

The triangles C E F and A E B have the sides C E and E F equal to the sides A E and E B (const.), and the angle C E F equal to A E B (XV), therefore the angles E C F and A are equal (IV), and therefore A C D is greater than A. In like manner it can be shown, that if A C be produced, the external angle B C G is greater than the angle B, and therefore that the angle A C D, which is equal to B C G (XV), is greater than the angle B.

(90)   Cor. 1.—Hence it follows, that each angle of a triangle is less than the supplement of either of the other angles (84). For the external angle is the supplement of the adjacent internal angle (XIII).

(91)   Cor. 2.—If one angle of a triangle be right or obtuse, the others must be acute. For the supplement of a right or obtuse angle is right or acute (82), and each of the other angles must be less than this supplemental, and must therefore be acute.

(92)   Cor. 3.—More than one perpendicular cannot be drawn from the same point to the same right line. For if two lines be supposed to be drawn, one of which is perpendicular, they will form a triangle having one right angle. The other angles must therefore be acute (91), and therefore the other line is not perpendicular.

(93)   Cor. 4.—If from any point a right line be drawn to a given right line, making with it an acute and obtuse angle, and from the same point a perpendicular be drawn, the perpendiculard must fall at the side of the acute angle. For otherwise a triangle would be formed having a right and an obtuse angle, which cannot be (91).

(94)   Cor. 5.—The The equal angles of an isosceles triangle must be both acute.

## Euclid, Elements of Geometry, Book I, Proposition 17

### Proposition 17[Heath's Edition]

In any triangle two angles taken together in any manner are less than two right angles.

Let ABC be a triangle;
I say that two angles of the triangle ABC taken together in any manner are less than two right angles.

For let BC be produced to D.

Then, since the angle ACD is an exterior angle of the triangle ABC,
it is greater than the interior and opposite angle ABC. [I. 16]
Let the angle ACB be added to each;
therefore the angles ACD, ACB are greater than the angles ABC, BCA. But the angles ACD, ACB are equal to two right angles. [I. 13]
Therefore the angles ABC, span class="geofig">BCA are less than two right angles.

Similarly we can prove that the angles BAC, ACB are also less than two right angles, and so are the angles CAB, ABC as well.

Therefore etc. Q.E.D.

Proposition XVII. Theorem. [Lardner's Edition]

 (95) Any two angles of a triangle (B A C) are together less than two right angles.

Produce any side B C, then the angle A C D is greater than either of the angles A or B (XVI), therefore A C B together with either A or B is less than the same angle A C B together with A C D; that is, less than two right angles (VIII). In the same manner, if C B be produced from the point B, it can be demonstrated that the angle A B C together the angle A is less than two right angles; therefore any two angles of the triangle are less than two right angles.

This proposition and the sixteenth are included in the thirty-second. which proves that the three angles are together equal to two right angles.

## Euclid, Elements of Geometry, Book I, Proposition 18

### Proposition 18[Heath's Edition]

In any triangle the greater side subtends the greater angle.

For let ABC be a triangle having the side AC greater than AB;
I say that the angle ABC is also greater than the angle BCA.

For, since AC is greater than AB, let AD be made equal to AB [I. 3] , and let BD be joined.

Then, since the angle ADB is an exterior angle of the triangle BCD,
it is greater than the interior and opposite angle DCB. [I. 16]

But the angle ADB is equal to the angle ABD,
since the side AB is equal to AD;
therefore the angle ABD is also greater than the angle ACB;
therefore the angle ABC is much greater than the angle ACB.

Therefore etc. Q.E.D.

Proposition XVIII. Theorem. [Lardner's Edition]

 (96) In any triangle (B A C) if one side (A C) be greater than another (A B), the angle opposite to the greater side is greater than the angle opposite to the less.

From the greater side A C cut off the part A D equal to the less (III), and conterminous with it, and join B D.

The triangle B A D being isosceles (V), the angles A B D and A D B are equal; but A D B is greater than the internal angle A C B (XVI): therefore A B D is greater than A C B, and therefore A B C is greater than A C B: but A B C is opposite the greater side A C, and A C B is opposite the less A B

This proposition might also be proved by producing the lesser side A B, and taking A E equal to the greater side. In this case the angle A E C is equal to A C E (V), and therefore greater than A C B. But A B C is greater than A E C (XVI), and therefore A B C is greater than A C B

## Euclid, Elements of Geometry, Book I, Proposition 19

### Proposition 19[Heath's Edition]

In any triangle the greater angle is subtended by the greater side.

Let ABC be a triangle having the angle ABC greater than the angle BCA;
I say that the side AC is also greater than the side AB.

For, if not, AC is either equal to AB or less.

Now AC is not equal to AB;
for then the angle ABC would also have been equal to the angle ACB; [I. 5]
but it is not;
therefore AC is not equal to AB.

Neither is AC less than AB,
for then the angle ABC would also have been less than the angle ACB; [I. 18]
but it is not;
therefore AC is not less than AB.

And it was proved that it is not equal either.
Therefore AC is greater than AB.

Therefore etc. Q.E.D.

Proposition XIX. Theorem. [Lardner's Edition]

 (97) If in any triangle (B A C) one angle (B) be greater than another (C), the side (A C) which is opposite the greater angle is greater than the side A B, which is opposie to the less.

For the side A C is either equal, or less, or greater than A B. It is not equal to A B, because the angle B would then be equal to C (V), which is contrary to the hypothesis.

It is not less than A B, because the angle B would then be less than C (XVIII), which is also contrary to the hypothesis.

Since therefore the side A C is neither equal to not less than A B, it is greater than it.

This proposition holds the same relation to the sixth, as the preceding does to the fifth. The four might be thus combined: one angle of a triangle is greater or less than another, or equal to it, according as the side opposed to the one is greater or less than, or equal to the side opposed to the other, and vice versa.

The student generally feels it difficult to remember which of the two, the eighteenth or nineteenth, is proved by construction, and which indirectly. By referring them to the fifth and sixth the difficulty will be removed.

## Euclid, Elements of Geometry, Book I, Proposition 20

### Proposition 20[Heath's Edition]

In any triangle two sides taken together in any manner are greater than the remaining one.

For let ABC be a triangle;
I say that in the triangle ABC two sides taken together in any manner are greater than the remaining one, namely

BA, AC greater than BC,
AB, BC greater than AC,
BC, CA greater than AB.

For let BA be drawn through to the point D, let DA be made equal to CA, and let DC be joined.

Then, since DA is equal to AC, the angle ADC is also equal to the angle ACD; [I. 5]
therefore the angle BCD is greater than the angle ADC. [C.N. 5]

And, since DCB is a triangle having the angle BCD greater than the angle BDC,
and the greater angle is subtended by the greater side, [I. 19]
therefore DB is greater than BC.

But DA is equal to AC;
therefore BA, AC are greater than BC.

Similarly we can prove that AB, BC are also greater than CA, and BC, CA than AB.

Therefore etc. Q.E.D.

Proposition XX. Theorem. [Lardner's Edition]

 (98) Any two sides (A B and A C) of a triangle (B A C) taken together, are greater than the third side (B C).

Let the side B A be produced, and let A D be cut off equal to A C (III), and let D C be drawn.

Since A D and A C are equal, the angles D and A C D are equal (V). Hence the angle B C D is greater than the angle D, and therefore the side B D in the triangle B C D is greater than B C (XIX). But B D is equal to B A and A C taken together, since A D was assumed equal to A C. Therefore B A and A C taken together are greater than B C.

This proposition is sometimes proved by bisecting the angle A. Let A E bisect it. The angle B E A is greater than E A C, and the angle C E A is greater than E A B (XVI); and since the parts of the angle A are equal, it follows, that each of the angles E is greater than each of the parts of A; and thence, by (XIX), it follows that B A is greater than B E, and A C greater than C E, and therefore that the sum of the former is greater than the sum of the latter.

The proposition might likewise be proved by drawing a perpendicular from the angle A on the side B C; but these methods seem inferior in clearness and brevity to that of Euclid.

Some geometers, among whom may be reckoned Archimedes, ridicule this proposition as being self evident, and contend that it should therefore be one of the axioms. That a truth is considered self evident is, however, not a sufficient reason why it should be adopted as a geometrical axiom (57).

(99)   It follows immediately from this proposition, that the difference of any two sides of a triangle is less than the remaining side For the sides A C and B C taken together are greater than A B; let the side A C be taken from both, and we shall have the side B C greater than the remainder upon taking A C from A B; that is, then the difference between A B and A C.

In this proof we assume something more than is expressed in the fifth axiom. For we take for granted, that if one quantity (a) be greater than another (b), and that equals be taken from both, the remainder of the former (a) will be grater than the remainder of the latter (b). This is a principle which is frequently used, though not directly expressed in the axiom (55).

## Euclid, Elements of Geometry, Book I, Proposition 21

### Proposition 21[Heath's Edition]

If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greater angle.

On BC, one of the sides of the triangle ABC, from its extremities, B, C, let the two straight lines BD, DC be constructed meeting within the triangle;
I say that BD, DC are less than the remaining two sides of the triangle BA, AC, but contain an angle BDC greater than the angle BAC.

For let BD be drawn through to E.

Then, since in any triangle two sides are greater than the remaining one, [I. 20]
therefore, in the triangle ABE, the two sides AB, AE are greater than BE.
Let EC be added to each,
therefore BA, AC are greater than BE, EC.

Again, since, in the triangle CED,
the two sides CE, ED are greater than CD,
let DB be added to each;
therefore CE, EB are greater than CD, DB.

But BA, AC, were proved greater than BE, EC;
therefore BA, AC are much greater than BD, DC.

Again, since in any triangle the exterior angle is greater than the interior and opposite angle, [I. 16]
therefore in the triangle CDE,
the exterior angle BDC is greater than the angle CED.

For the same reason, moreover, in the triangle ABE also, the exterior angle CEB is greater than the angle BAC.
But the angle BDC was proved greater than the angle CEB;
therefore the angle BDC is much greater than the angle BAC.

Therefore, etc. Q.E.D.

Proposition XXI. Theorem. [Lardner's Edition]

 (100) The sum of two right lines (D B and D C drawn to a point (D) within a triangle (B A C) from the extremities of any side (B C), is less than the sum of the other two sides of the triangle (A B and A C), but the lines contain a greater angle.

[INSERT PROOF.] Produce B D to E. The sum of the sides B A and A E of the triangle B A E is greater than the third side B E (XX); add E C to each, and the sum of the sides B A A C and is greater than the sum of B E and E C, but the sum of the sides D E and E C of the triangle D E C is greater than the third side D C (XX); and B D to each, and the sum of B E and E C is greater than the sume of B D and D C, but the sum of B A and A C is greater than that of B E and E C; therefore the sum of B A and A C is greater than that of B D and D C.

Because the external angle B D C is greater than the internal D E C (XVI), and for the same reason D E C is greater than A, the angle B D C is greater than the angle A.

By the thirty-second proposition it will follow, that the angle B D C exceeds the angle A by the sum of the angles A B D and A C D. For the angle B D C is equal to the sum of D E C and D C E; and, again, the angle D E C is equal to the sum of the angles A and A B E. Therefore the angle B D C is equal to the sum of A, and the angles A B D and A C D.

## Euclid, Elements of Geometry, Book I, Proposition 22

### Proposition 22[Heath's Edition]

Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one.

Let the three given straight lines be A, B, C, and of these let two taken together in any manner be greater than the remaining one,
namely

A, B greater than C,
A, C greater than B,
and
B, C greater than A;
thus it is required to construct a triangle out of straight lines equal to A, B, C.

Let there be set out a straight line DE, terminated at D but of infinite length in the direction of E, and let DF be made equal to A, FG equal to B, and GH equal to C.

With centre F and distance FD let the circle DKL be described;
again, with centre G and distance GH let the circle KLH be described;
and let KF, FG be joined;
I say that the triangle KFG has been constructed out of three straight lines equal to A, B, C.

For, since the point F is the centre of the circle DKL,
FD is equal to FK.
But FD is equal to A;
therefore KF is also equal to A.

Again, since the point G is the centre fo the circle LKH, GH is equal to GK.
But GH is equal to C;
therefore GK is equal to C.

And FG is also equal to B;
therefore the three straight lines KF, FG, GK are equal to the three straight lines A, B, C.

Therefore out of the three straight lines KF, FG, GK, which are equal to the three given straight lines A, B, C, the triangle KFG has been constucted. Q.E.F.

Proposition XXII. Problem. [Lardner's Edition]

 (101) Given three right lines (A, B and C) the sum of any two of which is greater than the third, to construct a triangle whose sides shall be respectively equal to the given lines.

### Solution.

From any point D draw the right line D E equal to one of the given lines A (II), and from the same point draw D G equal to another of the given lines B, and from the point Edraw E F equal to C. From the centre D with the radius D G describe a circle, and from the centre E with the radius E F describe another circle, and from a point K of intersection of these circles draw K D and K E.

### Demonstration.

It is evident, that the sides D E, D K and K E of the triangle D K E are equal to the given right lines A, B and C.

In this solution Euclid assumes that the two circles will have at least one point of intersection. To prove this, it is only necessary to show that a part of one of the circles will be within, and another part without the other (58).

Since D E and E K or E L are together greater than D K, it follows, that D L is greater than the radius of the circle K G, and therefore the point L is outside the circle. Also, since D K and E K are together greater than D E, if the equals E K and E H be taken from both, D H is less than D K, that is, D H is less than the radius of the circle, and therefore the point H is within it. Since the point H is within the circle and L without it, the one circle must intersect the other.

It is evident, that if the sum of the lines B and C were equal to the line A, the points H and K would coincide; for then the sum of D K and K E would equal D E. Also, if the sum of A and C were equal to B, the points K and L would coincide; for then D K would be equal to E K and D E, or to L D. It will hereafter appear, that in the former case the circles would touch externally, and in the latter internally.

If the line A were greater than the sum of B and C, it is easy to perceive that the circle would not meet, one being wholly outside the other; and if If the line B were greater than the sum of A and C, they would not meet, one being wholly within the other.

If the three right lines A B C be equal, this proposition becomes equivalent to the first, and the solution will be found to agree exactly with that of the first.

## Euclid, Elements of Geometry, Book I, Proposition 23

### Proposition 23[Heath's Edition]

On a given straight line and at a point on it to construct a rectilineal angle equal to a given rectilineal angle.

Let AB be the given straight line, A the point on it, and the angle DCE the given rectilineal angle;
thus it is required to construct on the given straight line AB, and at the point A on it, a rectilineal angle equal to the given rectilineal angle DCE.

On the straight lines CD, CE respectively let the points D, E be taken at random;
let DE be joined,
and out of three straight lines which are equal to the three straight lines CD, DE, CE let the triangle AFG be constructed in such a way that CD is equal to AF, CE to AG, and further DE to FG.

Then, since the two sides DC, CE are equal to the two sides FA, AG respectively,
and the base DE is equal to the base FG,
the angle DCE is equal to the angle FAG. [I. 8]

Therefore on the given straight line AB, and at the point A on it, the rectilineal angle FAG has been constructed equal to the given rectilineal angle DCE. Q.E.F.

Proposition XXIII. Problem. [Lardner's Edition]

 (102) At a given point (B) in a given right line (B E) to make an angle equal to a given angle (C).

### Solution.

In the sides of the given angle take any points D and F; join D F, and construct a triangle E B A which shall be equilateral with the triangle D C F, and whose sides A B and E B meeting at the given point B shall be equal to F C and D C of the given angle C (XXII). The angle E B A is equal to the given angle D C F.

### Demonstration.

For as the triangles D C F and E B A have all their sides respectively equal, the angles F C D and A B E opposite the equal sides D F and E A are equal (VIII).

It is evident that the eleventh proposition is a particular case of this

## Euclid, Elements of Geometry, Book I, Proposition 24

### Proposition 24[Heath's Edition]

If two triangles have the two sides equal to two sides respectively, but have the one of the angles contained by the equal straight lines greater than the other, they will also have the base greater than the base.

Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF, and let the angle at A be greater than the angle at D;
I say that the base BC is also greater than the base EF.

For, since the angle BAC is greater than the angle EDF, let there be constructed, on the straight line DE, and at the point D on it, the angle EDG equal to the angle BAC [I. 23] ; let DG be made equal to either of the two straight lines AC, DF, and let EG, FG be joined.

Then, since AB is equal to DE, and AC to DG, the two sides BA, AC are equal to the two sides ED, DG respectively;
and the angle BAC is equal to the angle EDG;
therefore the base BC is equal to the base EG. [I. 4]

Again, since DF is equal to DG,
the angle DGF is also equal to the angle DFG; [I. 5]
therefore the angle DFG is greater than the angle EGF.

Therefore the angle EFG is much greater than the angle EGF.

And, since EFG is a triangle having the angle EFG greater than the angle EGF,
and the greater angle is subtended by the greater side, [I. 19]
the side EG is also greater than EF.

But EG is equal to BC.
Therefore BC is also greater than EF.

Therefore etc. Q.E.D.

Proposition XXIV. Theorem. [Lardner's Edition]

 (103) If two triangles (E F D, B A C) have two sides of the one respectively equal to two sides of the other (F E to A B and F D to A C), and if one of the angles (B A C) contained by the equal sides be greater than the other (E F D), the side (B C) which is opposite to the greater angle is greater than the side (E D) which is opposite to the less angle.

From the point A draw the right line A G, making with the side A B, which is not the greater, an angle B A G equal to the angle E F D (XXIII). Make A G equal to F D (III), and draw B G and G C.

In the triangles B A G and E F D the sides B A and A G are equal respectively to E F and F D, and the included angles are equal (const.), and therefore B G is equal to E D. Also, since A G is equal to F D by const., and A C is equal to it by hyp., A G is equal to A C, therefore the triangles A C G and A G C are equal (V); but the angle B G C is greater than A G C, therefore greater than A C G, and therefore greater than B C G; then in the triangle B G C the angle B G C is greater than B C G, therefore the side B C is greater than B G (XIX), but B G is equal to E D, and therefore B C is greater than E D.

In this demonstration it is assumed by Euclid, that the points A and G will be on different sides of B C, or, in other words, that A H is less than A G or A C. This may be proved thus:—The side A C not being less than A B, the angle A B C cannot be less than the angle A C B (XVIII). But the angle A B C must be less than the angle A H C (XVI); therefore the angle A C B is less than A H C, and therefore A H less than A C or A G (XIX).

In the construction for this proposition Euclid has omitted the words ‘with the side which is not the greater.’ Without these it would not follow that the point G would fall below the base B C, and it would be necessary to give demonstrations for the cases in which the point G falls on, or above the base B C. On the other hand, if these words be inserted, it is necessary in order to give validity to the demonstration, to prove as above, that the point G falls below the base.

If the words ‘with the side not the greater’ be not inserted, the two omitted cases may be proved as follows:

If the point G fall on the base B C, it is evident that B G is less than B C (51).

If G fall above the base B C, let it be at G′. The sum of the lines B G′ and A G′ is less than the sum of A C and C B (XXI). The equals A C and A G′ being taken away, there will remain B G′ less than B C.

## Euclid, Elements of Geometry, Book I, Proposition 25

### Proposition 25[Heath's Edition]

If two triangles have the two sides equal to two sides respectively, but have the base greater than the base, they will also have the one of the angles contained by the equal straight lines greater than the other.

Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE and AC to DF; and let the base BC be greater than the base EF;
I say that the angle BAC is also greater than the angle EDF.

For, if not, it is either equal to it or less.

Now the angle BAC is not equal to the angle EDF;
for then the base BC would also have been equal to the base EF, [I. 4]
but it is not;
therefore the angle BAC is not equal to the angle DEF.

Neither again is the angle BAC less than the angle EDF;
for then the base BC would also have been less than the base EF, [I. 24]
but it is not;
therefore the angle BAC is not less than the angle EDF.

But it was proved that it is not equal either;
therefore the angle BAC is greater than the angle EDF.

Therefore, etc. Q.E.D.

Proposition XXV. Theorem. [Lardner's Edition]

 (104) If two triangles (B A C and E F D) have two sides of the one respectively equal to two of the other (B A to E F and A C to F D), and if the third side of the one (B C) be greater than the third side (E D) of the other, the angle (A) opposite the greater side is greater than the angle (F), which is opposite to the less.

The angle A is either equal to the angle F, or less than it, or greater than it.

It is not equal; for if it were, the side B C would be equal to the side E D (IV), which is contrary to the hypothesis.

It is not less; for if it were, the side B C would be less than the side E D (XXIV), which is contrary to the hypothesis.

Since therefore the angle A is neither equal to, nor less than F, it must be greater.

This proposition might be proved directly thus: On the greater side B C take B G equal to the lesser side E D, and on B G construct a triangle B H G equilateral with E F D. Join A H and produce H G to I.

The angle H will then be equal to the angle F.

1° Let B G be greater than B K.

Since B A and B H are equal, the angles B A H and B H A are equal (V). Also since H G is equal to A C, it is greater than A I, and therefore H I is greater than A I, and therefore the angle H A I is greater than the angle A H I (XVIII). Hence, if the equal triangles B H A and B A H be added to these, the angle B A C will be found greater than the angle B H G, which is equal to F.

2° If B G be not greater than If B K, it is evident that the angle H is less than the angle A.

The twenty-fourth and twenty-fifth propositions are analogous to the fourth and eighth, in the same manner as the eighteenth and nineteenth are to the fifth and sixth . The four might be announced together thus:

If two triangles have two sides of the one respectively equal to two sides of the other, the remaining side of the one will be greater or less than, or equal to the remaining side of the other, according as the angle opposed to it in the one is greater or less than, or equal to the angle opposed to it in the other, or vice versa.

In fact, these principles amount to this, that if two lines of given lengths be placed so that one pair of extremities coincide, and so that in their initial position the lesser line is placed upon the greater, the distance between the extremities will then be the difference of the lines. If they be opened as to form a gradually increasing angle, the line joining their extremities will gradually increase, until the angle they include becomes equal to two right angles, when they will be in one continued line, and the line joining their extremities is their sum. Thus the major and minor limits of this line is the sum and difference of the given lines. This evidently includes the twentieth proposition.

## Euclid, Elements of Geometry, Book I, Proposition 26

### Proposition 26[Heath's Edition]

If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle to the remaining angle.

Let ABC, DEF be two triangles having the two angles ABC, BCA equal to the two angles DEF, EFD respectively, namely the angle ABC to the angle DEF, and the angle BCA to the angle EFD; and let them also have one side equal to one side, first that adjoining the equal angles, namely BC to EF;
I say that they will also have the remaining sides equal to the remaining sides respectively, namely AB to DE and AC to DF, and the remaining angle to the remaining angle, namely the angle BAC to the angle EDF.

For if AB is unequal to DE, one of them is greater.

Let AB be greater, and let BG be made equal to DE;
and let GC be joined.

Then, since BG is equal to DE, and BC to EF,
the two sides GB, BC are equal to the two sides DE, EF respectively;
and the angle GBC is equal to the angle DEF;
therefore the base GC is equal to the base DF,
and the triangle GBC is equal to the triangle DEF,
and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4]
therefore the angle GCB is equal to the angle DFE.

But the angle DFE is by hypothesis equal to the angle BCA;
therefore the angle BCG is equal to the angle BCA,
the less to the greater: which is impossible.
Therefore AB is not unequal to DE,
and is therefore equal to it.

But BC is also equal to EF;
therefore the two sides AB, BC are equal to the two sides DE, EF respectively,
and the angle ABC is equal to the angle DEF;
therefore the base AC is equal to the base DF,
and the remaining angle BAC is equal to the remaining angle EDF. [I. 4]

Again, let sides subtending equal angles be equal, as AB to DE;
I say again that the remaining sides will be equal to the remaining sides, namely AC to DF and BC to EF, and further the remaining angle BAC is equal to the remaining angle EDF.

For if BC is unequal to EF, one of them is greater.

Let BC be greater, if possible, and let BH be made equal to EF; let AH be joined.

Then, since BH is equal to EF, and AB to DE,
the two sides AB, BH are equal to the two sides DE, EF respectively, and they contain equal angles;
therefore the base AH is equal to the base DF,
and the triangle ABH is equal to the triangle DEF,
and the remaining angles will be equal to the remaining angles, namely those which equal sides subtend; [I. 4]
therefore the angle BHA is equal to the angle EFD.

But the angle EFD is equal to the angle BCA;
therefore, in the triangle AHC, the exterior angle BHA is equal to the interior and opposite angle BCA:
which is impossible. [I. 16]
Therefore BC is not unequal to EF,
and is therefore equal to it.

But AB is also equal to DE;
therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and they contain equal angles;
therefore the base AC is equal to the base DF,
the triangle ABC equal to the triangle DEF,
and the remaining angle BAC equal to the remaining angle EDF. [I. 4]

Therefore etc. Q.E.D.

Proposition XXVI. Theorem. [Lardner's Edition]

 (105) If two triangles B A C, D E F have two angles of the one respectively equal to two angles of the other (B to D and C to F), and a side of the one equal to a side of the other similarly placed with respect to the equal angles, the remaining sides and angles are respectively equal to one another.

First let the equal sides be B C and D F, which lie between the equal angles; then the side B A is equal to the side D E.

For if it be possible, let one of them B A be greater than the other; make B G equal to D E, and join C G.

In the triangles G B C, E D F the sides G B, B C are respectively equal to the sides E D, D F (const.), and the angle B is equal to the angle D (hyp.), therefore the angles B C G and D F E are equal (IV); but the angle B C A is also equal to D F E (hyp.) therefore the angle B C G is equal to B C A (51), which is absurd: neither of the sides B A and D E therefore is greater than the other, and therefore they are equal, and also B C and D F are equal (IV), and the angles B and D; therefore the side A C is equal to the side E F, as also the angle A to the angle E (IV).

Next, let the equal sides be B A and D E, which are opposite to the equal angles C and F, and the sides B C and D F, shall also be equal.

For if it be possible, let one of them B C be greater than the other; make B G equal to D F, and join A G.

In the triangles A B G, E D F, the sides A B, B G are respectively equal to the sides E D, D F (const.), and the angle B is equal to the angle D (hyp.); therefore the angles A G B and E F D are equal (IV); but the angle C is also equal to E F D, therefore A G B and C are equal, which is absurd (XVI). Neither of the sides B C and D F is therefore greater than the other, and they are consequently equal. But B A and D E are also euqal, as also the angles B and D; therefore the side A C is equal to the side E F, and also the angle A to the angle E (IV).

It is evident that the triangles themselves are equal in every respect.

(106)   Cor. 1.—From this proposition and the principles previously established, it easily follows, that a line being drawn from the vertex of a triangle to the base, if any two of the following equalities be given (except the first two), the others may be inferred.

1° The equality of the sides of the triangle.

2° The equality of the angles at the base.

3° The equality of the angles under the line drawn, and the base.

4° The equality of the angles under the line drawn, and the sides.

5° The equality of the segments of the base.

Some of the cases of this investigation have already been proved (74), (75), (76). The others present no difficulty, except in the case where the fourth and fifth equalities are given to infer the others. This case may be proved as follows.

If the line A D which bisects the vertical angle (A) of a triangle also bisect the base B C, the triangle will be isosceles; for produce A D so that D E shall be equal to A D, and join E C. In the triangles D C E and A D B the angles vertically opposed at D are equal, and also the sides which contain them; therefore (IV) the angles B A D and D E C are equal, and also the sides A B and E C. But the angle B A D and D E C are equal, and also the sides A B and E C. But the angle B A D is equal to D A C (hyp.); and therefore D A C is equal to the angle E, therefore (VI) the sides A C and E C are equal. But A B and E C have already been proved equal, and therefore A B and A C are equal.

(107)   The twenty-sixth proposition furnishes the third criterion which has been established in the Elements for the equality of two triangles. It may be observed, that in a triangle there are six quantities which may enter into consideration, and in which two triangles may agree or differ; viz. the three sides and the three angles. We can in most cases infer the equality of two triangles in every respect, if they agree in any three of those six quantities which are independent of each other. To this, however, there are certain exceptions, as will appear by the following general investigation of the question.

When two triangles agree in three of the six quantities already mentioned, these three must be some of the six following combinations:

1° Two sides and the angle between them.

2° Two angles and the side between them.

3° Two sides, and the angle opposed to one of them.

4° Two angles, and the side opposed to one of them.

5° The three sides.

6° The three angles.

The first case has been established in the fourth, and the second and fourth in the twenty-sixth proposition. The fifth case has been established by the eighth, and in the sixth case the triangles are not necessarily equal. In this case, however, the three data are not independent, for it will appear by the thirty-second proposition, that any one angle of a triangle can be inferred from the other two.

The third is therefore the only case which remains to be investigated.

(108)   3° To determine under what circumstance two triangles having two sides equal each to each, and the angles opposed to one pair of equal sides equal, shall be equal in all respects. Let the sides A B and B C be equal to D E and E F, and the angle A be equal to the angle D. If the two angles B and E be equal, it is evident that the triangles are in every respect equal by (IV), and that C and F are equal. But if B and E be not equal, let one B be greater than the other E; and from B let a line B G be drawn, making the angle A B G equal to the angle E. In the triangles A B G and D E F, the angles A and A B G are equal respectively to D and E, and the side A B is equal to D E, therefore (XXVI) the triangles are in every respect equal; and the side B G is equal to E F, and the angle B G A equal to the angle F. But since E F is equal to B C, B G is equal to B C, and therefore (V) B G C is equal to B C G, and therefore C and B G A or F are supplemental.
(109)   Hence, if two triangles have two sides in the one respectively equal to two sides in the other, and the angles opposed to one pair of equal sides equal, the angles opposed to the other equal sides will be either equal or supplemental.

(110)   Hence it follows, that if two triangles have two sides respectively equal each to each, and the angles opposed to one pair of equal sides equal, the remaining angles will be equal, and therefore the triangles will be in every respect equal, if there be any circumstance from which it may be inferred that the angles opposed to the other pair of equal sides are of the same species.

(Angles are said to be of the same species when they are both acute, both obtuse, or both right).

For in this case, if they be not right they cannot be supplemental, and must therefore be equal (109), in which case the triangles will be in every respect equal, by (XXVI).

If they be both right, the triangles will be equal by (108); because in that case G and C being right angles, B G must coincide with B C, and the triangle B G A with B C A; but the triangle B G A is equal to E F D, therefore &c.

(111)   There are several circumstances which may determine the angles opposed to the other pair of equal sides to be of the same species, and therefore which will determine the equality of the triangles; amongst which are the following:

If one of the two angles opposed to the other pair of equal side be right; for a right angle is its own supplement.

If the angles which are given equal be obtuse or right; for then the other angles must be all acute (91), and therefore of the same species.

If the equal sides opposed to angles which are not given equal be less than the other sides, these angles must be both acute (XVIII).

In all these cases it may be inferred, that the triangles are in every respect equal.

It will appear by prop. 38, that if two triangles have two sides respectively equal, and the included angles supplemental, their areas are equal.

(The area of a figure is the quantity of surface within its perimeter).

(112)   If several right lines be drawn from a point to a given right line.

1° The shortest is that which is perpendicular to it.

2° Those equally inclined to the perpendicular are equal, and vice versa.

3° Those which meet the right line at equal distances from the perpendicular are equal, and vice versa.

4° Those which make greater angles with the perpendicular are greater, and vice versa.

5° Those which meet the line at greater distances from the perpendicular are greater, and vice versa.

6° More than two equal right lines cannot be drawn from the same point to the same right line.

The student will find no difficulty in establishing these principles.

(113)   If any number of isosceles triangles be constructed upon the same base, their vertices will be all placed upon the right line, which is perpendicular to the base, and passes through its middle point. This is a very obvious and simple example of a species of theorem which frequently occurs in geometrical investigations. This perpendicular is said to be the locus of the vertex of isosceles triangles standing on the same base.

## Euclid, Elements of Geometry, Book I, Proposition 27

### Proposition 27[Heath's Edition]

If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another.

For let the straight line EF falling on the two straight lines AB, CD make the alternate angles AEF, EFD equal to one another;
I say that AB is parallel to CD.

For, if not, AB, CD when produced will meet either in the direction of B, D or towards A, C.

Let them be produced and meet, in the direction of B, D, at G.

Then, in the triangle GEF, the exterior angle AEF is equal to the interior and opposite angle EFG:
which is impossible. [I. 16] Therefore AB, CD when produced will not meet in the direction of B, D.

Similarly it can be proved that neither will they meet towards A, C.

But straight lines which do not meet in either direction are parallel; [Def. 23]
therefore AB is parallel to CD.

Therefore etc. Q.E.D.

Proposition XXVII. Theorem. [Lardner's Edition]

 (114) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 28

### Proposition 28[Heath's Edition]

If a straight line falling on two straight lines make the exterior angle equal to the interior and opposite angle on the same side, or the interior angles on the same side equal to two right angles, the straight lines will be parallel to one another.

For let the straight line EF falling on the two straight lines AB, CD make the exterior angle EGB equal to the interior and opposite angle GHD, or the interior angles on the same side, namely BGH, GHD, equal to two right angles;
I say that AB is parallel to CD.

For, since the angle EGB is equal to the angle GHD,
while the angle EDB is equal to the angle AGH, [I. 15]
the angle AGH is also equal to the angle GHD;
and they are alternate;
therefore AB is parallel to CD. [I. 27]

Again, since the angles BGH, GHD are equal to two right angles, and the angles AGH, BGH are also equal to two right angles, [I. 13]
the angles AGH, BGH are equal to the angles BGH, GHD.

Let the angle BGH be subtracted from each;
therefore the remaining angle AGH is equal to the remaining angle GHD;
and they are alternate;
therefore AB is parallel to CD. [I. 27]

Therefore etc. Q.E.D.

Proposition XXVIII. Theorem. [Lardner's Edition]

 (115) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 29

### Proposition 29[Heath's Edition]

A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the interior angles on the same side equal to two right angles.

For let the straight line EF fall on the parallel straight lines AB, CD;
I say that it makes the alternate angles AGH, GHD equal, the exterior angle EGB equal to the interior and opposite angle GHD, and the interior angles on the same side, namely BGH, GHD, equal to two right angles.

For, if the angle AGH is unequal to the angle GHD, one of them is greater.

Let the angle AGH be greater.

Let the angle BGH be added to each;
therefore the angles AGH, BGH are greater than angles BGH, GHD.

But the angles AGH, BGH are equal to two right angles; [I. 13]
therefore the angles BGH, GHD are less than two right angles.

But straight lines produced indefinitely from angles less than two right angles meet [Post. 5] ; therefore AB, CD, if produced indefinitely, will meet;
but they do not meet, because they are by hypothesis parallel.

Therefore the angle AGH is not unequal to the angle GHD,
and is therefore equal to it.

Again, the angle AGH is equal to the angle EGB; [I. 15]
therefore the angle EGB is also equal to the angle GHD. [C.N. 1]
Let the angle BGH be added to each;
therefore the angles EGB, BGH are equal to the angles BGH, GHD. [C.N. 1]
But the angles EGB, BGH are equal to two right angles [I. 13]
therefore the angles BGH, GHD are also equal to two right angles.

Therefore etc. Q.E.D.

Proposition XXIX. Theorem. [Lardner's Edition]

 (116) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 30

### Proposition 30[Heath's Edition]

Straight lines parallel to the same straight line are also parallel to one another.

Let each of the straight lines AB, CD be parallel to EF;
I say that AB is also parallel to CD.

For let the straight line GK fall upon them.

Then, since the straight line GK has fallen on the parallel straight lines AB, EF,
the angle AGK is equal to the angle GHF. [I. 29]

Again, since the straight line GK has fallen on the parallel straight lines EF, CD,
the angle GHF is equal to the angle GKD. [I. 29]

But the angle AGK was also proved equal to the angle GHF;
therefore the angle AGK is also equal to the angle GKD; [C.N. 1]
and they are alternate.

Therefore AB is parallel to CD. Q.E.D.

Proposition XXX. Theorem. [Lardner's Edition]

 (121) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 31

### Proposition 31[Heath's Edition]

Through a given point to draw a straight line parallel to a given straight line.

Let A be the given point, and BC the given straight line;
thus it is required to draw through the point A a straight line parallel to the straight line BC.

Let a point D be taken at random on BC, and let AD be joined; on the straight line DA, and at the point A on it, let the angle DAE be constructed equal to the angle ADC [I. 23] ; and let the straight line AF be produced in a straight line with EA.

Then, since the straight line AD falling on the two straight lines BC, EF has made the alternate angles EAD, ADC equal to one another,
therefore EAF is parallel to BC. [I. 27]

Therefore through the given point A the straight line EAF has been drawn parallel to the given straight line BC. Q.E.F.

Proposition XXXI. Problem. [Lardner's Edition]

 (123) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 32

### Proposition 32[Heath's Edition]

In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of the triangle are equal to two right angles.

Let ABC be a triangle, and let one side of it BC be produced to D;
I say that the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC, and the three interior angles of the triangle ABC, BCA, CAB are equal to two right angles.

For let CE be drawn through the point C parallel to the straight line AB. [I. 31]

Then since AB is parallel to CE,
and AC has fallen upon them,
the alternate angles BAC, ACE are equal to one another. [I. 29]

Again, since AB is parallel to CE,
and the straight line BD has fallen upon them,
the exterior angle ECD is equal to the interior and opposite angle ABC. [I. 29]

But the angle ACE was also proved equal to the angle BAC;
therefore the whole angle ACD is equal to the two interior and opposite angles BAC, ABC.

Let the angle ACB be added to each;
therefore the angles ACD, ACB are equal to the three angles ABC, BCA, CAB.

But the angles ACD, ACB are equal to two right angles; [I. 13] therefore the angles ABC, BCA, CAB are also equal to two right angles.

Therefore etc. Q.E.D.

Proposition XXXII. Theorem. [Lardner's Edition]

 (124) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 33

### Proposition 33[Heath's Edition]

The straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are themselves also equal and parallel.

Let AB, CD be equal and parallel, and let the straight lines AC, BD join them (at the extremities which are) in the same directions (respectively);
I say that AC, BD are also equal and parallel.

Let BC be joined.

Then, since AB is parallel to CD, and BC has fallen upon them,
the alternate angles ABC, BCD are equal to one another. [I. 29]

And, since AB is equal to CD,
and BC is common,
the two sides AB, BC are equal to the two sides DC, CB;
and the angle ABC is equal to the angle BCD;
therefore the base AC is equal to the base BD,
and the triangle ABC is equal to the triangle DCB,
and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4]
therefore the angle ACB is equal to the angle CBD.

And, since the straight line BC falling on the two straight lines AC, BD has made the alternate angles equal to one another,
AC is parallel to BD. [I. 27]

And it was proved equal to it.

Therefore etc. Q.E.D.

Proposition XXXIII. Theorem. [Lardner's Edition]

 (150) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 34

### Proposition 34[Heath's Edition]

In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.

Let ACDB be a parallelogrammic area, and BC its diameter;
I say that the opposite sides and angles of the parallelogram ACDB are equal to one another, and the diameter BC bisects it.

For, since AB is parallel to CD,
and the straight line BC has fallen upon them,
the alternate angles ABC, BCD are equal to one another. [I. 29]

Again, since AC is parallel to BD,
and BC has fallen upon them,
the alternate angles ACB, CBD are equal to one another. [I. 29]

Therefore ABC, DCB are two triangles having the two angles ABC, BCA equal to the two angles DCB, CBD respectively, and one side equal to one side, namely that adjoining the equal angles and common to both of them, BC;
therefore they will also have the remaining sides equal to the remaining sides respectively, and the remaining angle to the remaining angle; [I. 26]
therefore the side AB is equal to CD,
and AC to BD,
and further the angle BAC is equal to the angle CDB.

And, since the angle ABC is equal to the angle BCD,
and the angle CBD to the angle ACB,
the whole angle ABD is equal to the whole angle ACD. [C.N. 2]
And the angle BAC was also proved equal to the angle CDB.

Therefore in parallelogrammic areas the opposite sides and angles are equal to one another.

I say, next, that the diameter also bisects the areas.

For since AB is equal to CD,
and BC is common,
the two sides AB, BC are equal to the two sides DC, CB respectively;
and the angle ABC is equal to the angle BCD;
therefore the base AC is also equal to DB,
and the triangle ABC is equal to the triangle DCB. [I. 4]

Therefore the diameter BC bisects the parallelogram ACDB. Q.E.D.

Proposition XXXIV. Theorem. [Lardner's Edition]

 (151) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 35

### Proposition 35[Heath's Edition]

Parallelograms which are one the same base and in the same parallels are equal to one another.

Let ABCD, EBCF be parallelograms on the same base BC and in the same parallels AF, BC;
I say that ABCD is equal to the parallelogram EBCF.

For, since ABCD is a parallelogram,
AD is equal to BC. [I. 34]

For the same reason also
EF is equal to BC,
so that AD is also equal to EF; [C.N. 1]
and DE is common;
therefore the whole AE is equal to the whole DF. [C.N. 2]

But AB is also equal to DC; [I. 34]
therefore the two sides EA, AB are equal to the two sides FD, DC respectively,
and the angle FDC is equal to the angle EAB, the exterior to the interior; [I. 29]
therefore the base EB is equal to the base FC,
and the triangle EAB will be equal to the triangle FDC. [I. 4]

Let DGE be subtracted from each;
therefore the trapazium ABGD which remains is equal to the trapezium EGCF which remains. [C.N. 3]

Let the triangle GBC be added to each;
therefore the thole parallelogram ABCD is equal to the whole parallelogram EBCF. [C.N. 2]

Therefore, etc. Q.E.D.

Proposition XXXV. Theorem. [Lardner's Edition]

 (165) [INSERT STATEMENT.]

[INSERT PROOF.] [ADJUST FIGURES FROM HERE ONWARDS.]
[CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 36

### Proposition 36[Heath's Edition]

Parallelograms which are on equal bases and in the same parallels are equal to one another.

Let ABCD, EFGH be parallelograms which are on equal bases BC, FG and in the same parallels AH, BG;
I say that the parallelogram ABCD is equal to EFGH.

For let BE, CH be joined.

Then, since BC is equal to FG
while FG is equal to EH,
BC is also equal to EH. [C.N. 1]

But they are also parallel.
And EB, HC join them;
but straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are equal and parallel. [I. 33]
Therefore EBCH is a parallelogram. [I. 34]
And it is equal to ABCD;
for it has the same base BC with it, and is in the same parallels BC, AH with it. [I. 35]

For the same reason also EFGH is equal to the same; EBCH [I. 35]
so that the parallelogram ABCD is also equal to EFGH. [C.N. 1]

Therefore etc. Q.E.D.

Proposition XXXVI. Theorem. [Lardner's Edition]

 (166) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 37

### Proposition 37[Heath's Edition]

Triangles which are on the same base and in the same parallels are equal to one another.

Let ABC, DBC be triangles on the same base BC and in the same parallels AD, BC;
I say that the triangle ABC is equal to the triangle DBC.

Let AD be produced in both directions to E, F;
through B let BE be drawn parallel to CA, [I. 31]
and through C let CF be drawn parallel to BD. [I. 31]

Then each of the figures EBCA, DBCF is a parallelogram; and they are equal,
for they are on the same base BC and in the same parallels BC, EF. [I. 35]

Moreover the triangle ABC is half of the parallelogram EBCA; for the diameter AB bisects it. [I. 34]

And the triangle DBC is half of the parallelogram DBCF; for the diameter DC bisects it. [I. 34]

[But halves of equal things are equal to one another.]

Therefore the triangle ABC is equal to the triangle DBC.

Therefore, etc. Q.E.D.

Proposition XXXVII. Theorem. [Lardner's Edition]

 (168) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 38

### Proposition 38[Heath's Edition]

Triangles which are on equal bases and in the same parallels are equal to one another.

Let ABC, DEF be triangles on equal bases BC, EF and in the same parallels BF, AD;
I say that the triangle ABC is equal to the triangle DEF.

For let AD be produced in both directions to G, H;
through B let BG be drawn parallel to CA, [I. 31]
and through F let FH be drawn parallel to DE.

Then each of the figures GBCA, DEFH is a parallelogram;
and GBCA is equal to DEFH;
for they are on equal bases BC, EF and in the same parallels BF, GH. [I. 36]

Moreover the triangle ABC is half of the parallelogram GBCA; for the diameter AB bisects it. [I. 34]

And the triangle ABC is half of the parallelogram DEFH; for the diameter DF bisects it. [I. 34]

[But the halves of equal things are equal to one another.]

Therefore the triangle ABC is equal to the triangle DEF.

Therefore etc. Q.E.D.

Proposition XXXVIII. Theorem. [Lardner's Edition]

 (169) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 39

### Proposition 39[Heath's Edition]

Equal triangles which are on the same base and on the same side are also in the same parallels.

Let ABC, DBC be equal triangles which are on the same base BC and on the same side of it;
[I say that they are also in the same parallels.] And [For] let AD be joined;
I say that AD is parallel to BC.

For, if not, let AE be drawn through the point A parallel to the straight line BC, [I. 31]
and let EC be joined.

Therefore the triangle ABC is equal to the triangle EBC;
for it is on the same base BC with it and in the same parallels. [I. 37]

But ABC is equal to DBC;
therefore DBC is also equal to EBC, [C.N. 1]
the greater to the less: which is impossible.
Therefore AE is not parallel to BC.

Similarly we can prove that neither is any other straight line except AD;
therefore AD is parallel to BC.

Therefore etc. Q.E.D.

Proposition XXXIX. Theorem. [Lardner's Edition]

 (172) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 40

### Proposition 40[Heath's Edition]

Equal triangles which are on equal bases and on the same side are also in the same parallels.

Let ABC, CDE be equal triangles on equal bases BC, CE and on the same side.

I say that they are also in the same parallels.

I say that AD is parallel to BE.

For, if not, let AF be drawn through A parallel to BE [I. 31] , and let FE be joined.

Therefore the triangle ABC is equal to the triangle FCE;
for they are on equal bases BC, CE and in the same parallels BE, AF.

But the triangle ABC is equal to the triangle DCE;
therefore the triangle DCE is also equal to the triangle FCE, [C.N. 1]
the greater to the less: which is impossible.

Therefore AF is not parallel to BE.

Similarly we can prove that neither is any other straight line except AD;
therefore AD is parallel to BE.

Therefore etc. Q.E.D.

Proposition XL. Theorem. [Lardner's Edition]

 (173) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 41

### Proposition 41[Heath's Edition]

If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle.

For let the parallelogram ABCD have the same base BC with the triangle EBC, and let it be in the same parallels BC, AE;
I say that the parallelogram ABCD is double of the triangle BEC.

For let AC be joined.
Then the triangle ABC is equal to the triangle EBC;
for it is on the same base BC with it and in the same parallels BC, AE. [I. 37]

But the parallelogram ABCD is double of the triangle ABC;
for the diameter AC bisects it; [I. 34]
so that the parallelogram ABCD is also double of the triangle EBC.

Therefore, etc. Q.E.D.

Proposition XLI. Theorem. [Lardner's Edition]

 (190) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 42

### Proposition 42[Heath's Edition]

To construct, in a given rectilineal angle, a parallelogram equal to a given triangle.

Let ABC be the given triangle, and D the given rectilineal angle;
thus it is required to construct in the rectilinear angle D a parallelogram equal to the triangle ABC.

Let BC be bisected at E, and let AE be joined;
on the straight line EC, and at the point E on it, let the angle CEF be constructed equal to the angle D; [I. 23]
through A let AG be drawn parallel to EC, [I. 31]
and through C let CG be drawn parallel to EF.

Then FECG is a parallelogram.

And, since BE is equal to EC,
the triangle ABE is also equal to the triangle AEC,
for they are on equal bases BE, EC and in the same parallels BC, AG;
therefore the triangle ABC is double of the triangle AEC.

But the parallelogram FECG is also double of the triangle AEC, for it has the same base with it and is in the same parallels with it; [I. 41]
therefore the parallelogram FECG is equal to the triangle ABC.
And it has the angle CEF equal to the given angle D.

Therefore the parallelogram FECG has been constructed equal to the given triangle ABC, in the angle CEF which is equal to D. Q.E.F.

Proposition XLII. Problem. [Lardner's Edition]

 (194) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 43

### Proposition 43[Heath's Edition]

In any parallelogram the complements of the parallelograms about the diameter are equal to one another.

Let ABCD be a parallelogram, and AC its diameter;
and about AC let EH, FG be parallelograms, and BK, KD the so-called complements;
I say that the complement BK is equal to the complement KD.

For since ABCD is a parallelogram, and AC its diameter,
the triangle ABC is equal to the triangle ACD. [I. 34]

Again, since EH is a parallelogram, and AK is its diameter,
the triangle AEK is equal to the triangle AHK.

For the same reason
the triangle KFC is also equal to KGC.

Now, since the triangle AEK is equal to the triangle AHK,
and KFC to KGC,
the triangle AEK together with KGC is equal to the triangle AHK together with KFC. [C.N. 2]

And the whole triangle ABC is also equal to the whole ADC;
therefore the complement BK which remains is equal to the complement KD which remains. [C.N. 3]

Therefore etc. Q.E.D.

Proposition XLIII. Theorem. [Lardner's Edition]

 (195) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 44

### Proposition 44[Heath's Edition]

To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.

Let AB be the given straight line, C the given triangle and D the given rectilineal angle;
thus it is required to apply to the given straight line AB, in an angle equal to the angle D, a parallelogram equal to the given triangle C.

Let the parallelogram BEFG be constructed equal to the triangle C, in the angle EBG which is equal to D [I. 42] ; let it be placed so that BE is in a straight line with AB; let FG be drawn through to H, and let AH be drawn through A parallel to either BG or EF [I. 31]

Let HB be joined.

Then, since the straight line HF falls upon the parallels AH, EF,
the angles AHF, HFE are equal to two right angles. [I. 29]
Therefore the angles BHG, GFE are less than two right angles;
and straight lines produced indefinitely from angles less than two right angles meet; [Post. 5]
therefore HB, FE, when produced, will meet.

Let them be produced and meet at K; through the point K let KL be drawn parallel to either EA or FH, [I. 31]
and let HA, GB be produced to the points L, M.

Then HLKF is a parallelogram,
HK is its diameter, and AG, ME are parallelograms, and LB, BF the so-called complements, about HK;
therefore LB is equal to BF. [I. 43]

But BF is equal to the triangle C;
therefore LB is also equal to C. [C.N. 1]

And, since the angle GBE is equal to the angle ABM, [I. 15]
while the angle GBE is equal to D,
the angle ABM is also equal to the angle D.

Therefore the parallelogram LB equal to the given triangle C has been applied to the given straight line AB, in the angle ABM which is equal to D. Q.E.F.

Proposition XLIV. Problem. [Lardner's Edition]

 (198) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 45

### Proposition 45[Heath's Edition]

To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure.

Let ABCD be the given rectilineal figure and E the given rectilineal angle;
thus it is required to construct, in the given angle E, a parallelogram equal to the rectilineal figure ABCD.

Let DB be joined, and let the parallelogram FH be constructed equal to the triangle ABD, in the angle HKF which is equal to E;
let the parallelogram GM equal to the trangle DBC be applied to the straight line GH, in the angle GHM which is equal to E.

Then, since the angle E is equal to each of the angles HKF, GHM,
the angle HKF is also equal to the angle GHM. [C.N. 1]

Let the angle KHG be added to each;
therefore the angles KFH, KHG are equal to the angles KHG, GHM.

But the angles FKH, KHG are equal to two right angles; [I. 29]
thereore the angles KHG, GHM are also equal to two right angles.

Thus, with a straight line GH, and at the point H on it, two straight lines KH, HM not lying on the same side make the adjacent angles equal to two right angles;
therefore HK is in a straight line with HM [I. 14]

And, since the straight line HG falls upon the parallels KM, FG, the alternate angles MHG, HGF are equal to one another. [I. 29]

Let the angle HGL be added to each;
therefore the angles MHG, HGL are equal to the angles HGF, HGL. [C.N. 2]

But the angles MHG, HGL are equal to two right angles; [I. 29]
therefore the angles HGF, HGL are also equal to two right angles. [C.N. 1]
Therefore FG is in a straight line with GL [I. 14]

And, since FK is equal and parallel to HG, [I. 34]
and HG to ML also,
KF is also equal and parallel to ML; [C.N. 1; ] [I. 30]
and the straight lines KM, FL join them (at their extremities); therefore KM, FL are also equal and parallel. [I. 33]
Therefore KFLM is a parallelogram.

And, since the triangle ABD is equal to the parallelogram FH,
and DBC to GM,
the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM.

Therefore the parallelogram KFLM has been constructed equal to the given rectilineal figure ABCD, in the angle FKM which is equal to the given angle E. Q.E.F.

Proposition XLV. Problem. [Lardner's Edition]

 (199) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 46

### Proposition 46[Heath's Edition]

On a given straight line to describe a square.

Let AB be the given straight line;
thus it is required to describe a square on the straight line AB.

Let AC be drawn at right angles to the straight line AB from the point A on it, [I. 11] and let AD be made equal to AB;
through the point D let DE be drawn parallel to AB,
and through the point B let BE be drawn parallel to AD. [I. 31]

therefore AB is equal to DE, and AD to BE [I. 34] . But AB is equal to AD;
therefore the four straight lines BA, AD, DE, EB are equal to one another;
therefore the parallelogram ADEB is equilateral.

I say next that it is also right-angled.

For, since the straight line AD falls upon the parallels AB, DE,
the angles BAD, ADE are equal to two right angles. [I. 29]
But the angle BAD is right;
therefore the angle ADE is also right.
And in parallelogrammic areas the opposite sides and angles are equal to one another; [I. 34]
therefore each of the opposite angles ABE, BED is also right.

And it was also proved equilateral.

Therefore it is a square; and it is described on the straight line AB. Q.E.F.

Proposition XLVI. Problem. [Lardner's Edition]

 (201) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 47

### Proposition 47[Heath's Edition]

In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.

Let ABC be a right-angled triangle having the angle BAC right;
I say that the square on BC is equal to the squares on BA, AC.

For let there be described on BC the square BDEC,
and on BA, AC the squares GB, HC; [I. 46]
through A let AL be drawn parallel to either BD or CE, and let AD, FC be joined.

Then, since each of the angles BAC, BAG is right, if follows that with a straight line BA, and at the point A on it, the two straight lines AC, AG not lying on the same side make the adjacent angles equal to two right angles;
therefore CA is in a straight line with AG. [I. 14]

For the same reason
BA is also in a straight line with AH.

And, since the angle DBC is equal to the angle FBA: for each is right:
let the angle ABC be added to each;
therefore the whole angle DBA is equal to the whole angle FBC. [C.N. 2]

And, since DB is equal to BC, and FB to BA,
the two sides AB, BD are equal to the two sides FB, BC respectively;
and the angle ABD is equal to the angle FBC;
therefore the base AD is equal to the base FC,
and the triangle ABD is equal to the triangle FBC. [I. 4]

Now the parallelogram BL is double of the triangle ABD, for they have the same base BD and are in the same parallels BD, AL. [I. 41]
And the square GB is double of the triangle FBC,
for they again have the same base FB and are in the same parallels. FB, GC [I. 41]
[But the doubles of equals are equal to one another.]
Therefore the parallelogram BL is also equal to the square GB.

Similarly, if AE, BK be joined,
the parallelogram CL can also be proved equal to the square HC;
therefore the whole square BDEC is equal to the two squares GB, HC. [C.N. 2]
And the square BDEC is described on BC,
and the squares GB, HC on BA, AC.

Therefore the square on the side BC is equal to the squares on the sides BA, AC.

Therefore etc. Q.E.D.

Proposition XLVII. Theorem. [Lardner's Edition]

 (202) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]

## Euclid, Elements of Geometry, Book I, Proposition 48

### Proposition 48[Heath's Edition]

If in a triangle the square on one of the sides be equal to the squares on the remaining two sides of the triangle, the angle contained by the remaining two sides of the triangle is right.

For in the triangle ABC let the square on one side BC be equal to the squares on the sides BA, AC;
I say that the angle BAC is right.

For let AD be drawn from the point A at right angles to the straight line AC, let AD be made equal to BA, and let DC be joined.

Since DA is equal to AB,
the square on DA is also equal to the square on AB.

therefore the squares on DA, AC are equal to the squares on BA, AC.

But the square on DC is equal to the squares on DA, AC, for the angle DAC is right; [I. 47]
and the square on BC is equal to the squares on BA, AC, for this is the hypothesis;
therefore the square on DC is equal to the square on BC,
so that the side DC is also equal to BC.

And since DA is equal to AB,
and AC is common,
the two sides DA, AC are equal to the two sides BA, AC;
and the base DC is equal to the base BC;
therefore the angle DAC is equal to the angle BAC. [I. 8]
But the angle DAC is right;
therefore the angle BAC is also right.

Therefore etc. Q.E.D.

Proposition XLVIII. Theorem. [Lardner's Edition]

 (216) [INSERT STATEMENT.]

[INSERT PROOF.] [CONTINUE PROOF.]