Definitions [Heath's Edition]

Definitions [Lardner's Edition]

Postulates [Heath's Edition]Let the following be postulated:

Postulates [Lardner's Edition]

Common Notions [Heath's Edition]

Axioms [Lardner's Edition]

Propositions 

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]
Now since the point A is the
centre of the
circle CDB, 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;
And things which are equal to the same thing are also
equal to one another;
[C.N. 1]
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. 
Proposition I. Problem. [Lardner's Edition]
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). 

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]
Let the straight lines
AE, BF
be produced in a straight
line with DA,
DB;
[Post. 2]
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;
But BC was also proved
equal to BG;
And things which are equal to the same thing are also equal
to one another;
[C.N. 1]
Therefore at the given point A
the straight line AL is
placed equal to the given straight
line BC. 
Proposition II. Problem. [Lardner's Edition]
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]
Now, since the point A is the
centre of the
circle DEF, 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. 
Proposition III. Problem. [Lardner's Edition]
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,
Again, AB coinciding with
DE,
the straight line AC will
also coincide with DF, because the
angle BAC is equal to
the angle EDF;
But B also coincided with
E; [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 the remaining angles will also coincide with the remaining
angles and will be equal to them,
Therefore etc. 
Proposition IV. Theorem. [Lardner's Edition]
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; 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;
Then, since AF is
equal to AG and
AB
to AC,
Therefore the base FC
is equal to the base GB, And, since [I. 4]
Again, let sides subtending equal angles be equal,
as AB
to DE; 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,
But the angle EFD is equal to
the angle BCA;
But AB is also
equal to DE; Therefore etc. Q.E.D. 
Proposition V. Theorem. [Lardner's Edition]
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.


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; 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;
Then, since DB is equal
to AC, and
BC is common,
Therefore AB is not unequal
to AC; Therefore etc. Q.E.D. 
Proposition VI. Theorem. [Lardner's Edition]
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.
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 selfevident, 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,
Again, since CD is
equal to DB, Therefore etc. Q.E.D. 
Proposition VII. Theorem. [Lardner's Edition]
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;
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,
Then, BC coinciding with
EF, 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; If therefore etc. Q.E.D. 
Proposition VIII. Theorem. [Lardner's Edition]
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.
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; I say that the angle BAC has been bisected by the straight line AF.
For, since AD is equal to
AE, Therefore the given rectilineal angle BAC has been bisected by the straight line AF. Q.E.F. 
Proposition IX. Problem. [Lardner's Edition]
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]
For, since AC is equal to
CB, and CD
is common, Therefore the given finite straight line AB has been bisected at D. Q.E.F. 
Proposition X. Problem. [Lardner's Edition]
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;
For, since DC is equal to
CE,
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 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]
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.
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;
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]
For, since GH is equal
to HE, 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]
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.
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.
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)
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.
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.
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)


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; 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]
Then, since the angle CBE is equal
to the two angles CBA,
ABE,
Again, since the angle DBA
is equal to the two angles
DBE,
EBA,
But the angles CBE,
EBD were also proved equal to the
same three angles; Therefore etc. Q.E.D. 
Proposition XIII. Theorem. [Lardner's Edition]
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; 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,
Let the angle CBA be subtracted
from each;
Similarly we can prove that neither is any other straight line
except BD. Therefore etc. Q.E.D. 
Proposition XIV. Theorem. [Lardner's Edition]
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;
For, since the straight line AE stands
on the straight line CD,
making the angles CEA,
AED,
Again, since the straight line DE
stands on the straight line
AB, making the
angles AED,
DEB,
But the angles CEA,
AED were also proved equal
to two right angles; 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]
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;
Let AC be bisected
at E
[I. 10]
,
and let BE be joined
and produced in a straight line
to F;
Then, since AE is equal
to EC, and
BE to
EF,
But the angle ECD is greater
than the angle ECF;
[C.N. 5]
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]
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; For let BC be produced to D.
Then, since the angle ACD is an
exterior angle of the
triangle ABC, 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]
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 thirtysecond. 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; 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,
But the angle ADB is equal to the
angle ABD, Therefore etc. Q.E.D. 
Proposition XVIII. Theorem. [Lardner's Edition]
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 

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; For, if not, AC is either equal to AB or less.
Now AC is not equal
to AB;
Neither is AC less than
AB,
And it was proved that it is not equal either. Therefore etc. Q.E.D. 
Proposition XIX. Theorem. [Lardner's Edition]
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; BA, AC greater than BC, 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]
And, since DCB is a triangle
having the angle BCD greater
than the angle BDC,
But DA is equal
to AC; 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]
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; For let BD be drawn through to E.
Then, since in any triangle two sides are greater than
the remaining one,
[I. 20]
Again, since, in the
triangle CED,
But BA, AC,
were proved greater than
BE,
EC;
Again, since in any triangle the exterior angle is
greater than the interior and opposite angle,
[I. 16]
For the same reason, moreover, in the
triangle ABE also,
the exterior angle CEB is greater
than the angle BAC. Therefore, etc. Q.E.D. 
Proposition XXI. Theorem. [Lardner's Edition]
[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 thirtysecond 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, A, B greater than C,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;
For, since the point F
is the centre of the circle
DKL,
Again, since the point G
is the centre fo the circle LKH,
GH is equal
to GK.
And FG is also
equal to B; 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]
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;
On the straight lines CD,
CE respectively let the points
D, E
be taken at random;
Then, since the two sides DC,
CE are equal to the two
sides FA,
AG respectively, 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]
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; 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;
Again, since DF is
equal to DG, 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,
But EG is equal to
BC. Therefore etc. Q.E.D. 
Proposition XXIV. Theorem. [Lardner's Edition]
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; For, if not, it is either equal to it or less.
Now the angle BAC is not equal to
the angle EDF;
Neither again is the angle BAC
less than the angle EDF;
But it was proved that it is not equal either; Therefore, etc. Q.E.D. 
Proposition XXV. Theorem. [Lardner's Edition]
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 twentyfourth and twentyfifth 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; For if AB is unequal to DE, one of them is greater.
Let AB be greater, and let
BG be made equal to
DE;
Then, since BG
is equal to DE,
and BC to
EF,
But the angle DFE
is by hypothesis equal to the angle
BCA;
But BC is also equal to
EF;
Again, let sides subtending equal angles be
equal, as AB to
DE; 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,
But the angle EFD is equal to
the angle BCA;
But AB is also equal to
DE; Therefore etc. Q.E.D. 
Proposition XXVI. Theorem. [Lardner's Edition]
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 twentysixth 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 twentysixth 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 thirtysecond 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. ^{★}★^{★} (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; 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: 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 etc. Q.E.D. 
Proposition XXVII. Theorem. [Lardner's Edition]
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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;
For, since the angle EGB
is equal to the
angle GHD,
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]
Let the angle BGH be subtracted from
each; Therefore etc. Q.E.D. 
Proposition XXVIII. Theorem. [Lardner's Edition]
[INSERT PROOF.] [CONTINUE PROOF.] [ADD COMMENTARY.] 

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; 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;
But the angles AGH,
BGH are equal to two right angles;
[I. 13]
But straight lines produced indefinitely from angles less than
two right angles meet
[Post. 5]
;
therefore
AB, CD,
if produced indefinitely, will meet;
Therefore the angle AGH is not
unequal to the
angle GHD,
Again, the angle AGH is equal to
the angle EGB;
[I. 15]
Therefore etc. Q.E.D. 
Proposition XXIX. Theorem. [Lardner's Edition]
[INSERT PROOF.] [CONTINUE PROOF.] [ADD COMMENTARY.] 

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; For let the straight line GK fall upon them.
Then, since the straight line GK
has fallen on the parallel straight lines
AB,
EF,
Again, since the straight line GK
has fallen on the parallel straight lines
EF,
CD,
But the angle AGK was also proved
equal to the angle
GHF; Therefore AB is parallel to CD. Q.E.D. 
Proposition XXX. Theorem. [Lardner's Edition]
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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; 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 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]
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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; For let CE be drawn through the point C parallel to the straight line AB. [I. 31]
Then since AB is parallel
to CE,
Again, since AB is
parallel to CE,
But the angle ACE was also proved
equal to the angle BAC;
Let the angle ACB be added
to each; 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]
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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); Let BC be joined.
Then, since AB is parallel to
CD, and
BC has fallen
upon them,
And, since AB is equal to
CD,
And, since the straight line BC
falling on the two straight lines
AC, BD
has made the alternate angles equal to one another, And it was proved equal to it. Therefore etc. Q.E.D. 
Proposition XXXIII. Theorem. [Lardner's Edition]
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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;
For, since AB is
parallel to CD,
Again, since AC is parallel to
BD,
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;
And, since the angle ABC
is equal to the
angle BCD, 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, Therefore the diameter BC bisects the parallelogram ACDB. Q.E.D. 
Proposition XXXIV. Theorem. [Lardner's Edition]
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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;
For, since ABCD is a
parallelogram,
For the same reason also
But AB is also equal to
DC;
[I. 34]
Let DGE be subtracted
from each;
Let the triangle GBC
be added to each; Therefore, etc. Q.E.D. 
Proposition XXXV. Theorem. [Lardner's Edition]
[INSERT PROOF.]
[ADJUST FIGURES FROM HERE ONWARDS.] [ADD COMMENTARY.] 

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; For let BE, CH be joined.
Then, since BC is equal to
FG
But they are also parallel.
For the same reason also EFGH
is equal to the same;
EBCH
[I. 35]
Therefore etc. Q.E.D. 
Proposition XXXVI. Theorem. [Lardner's Edition]
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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;
Let AD be produced in both directions
to E,
F;
Then each of the figures EBCA,
DBCF is a parallelogram;
and they are equal, 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]
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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;
For let AD be produced in
both directions to
G,
H;
Then each of the figures
GBCA, DEFH
is a parallelogram; 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]
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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;
For, if not, let AE be drawn through
the point A
parallel to the straight line BC,
[I. 31]
Therefore the triangle ABC is equal
to the triangle EBC;
But ABC is equal to
DBC;
Similarly we can prove that neither is any other straight line except
AD; Therefore etc. Q.E.D. 
Proposition XXXIX. Theorem. [Lardner's Edition]
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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.
For let AD be joined; 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;
But the triangle ABC is equal to
the triangle DCE; Therefore AF is not parallel to BE.
Similarly we can prove that neither is any other straight line
except AD; Therefore etc. Q.E.D. 
Proposition XL. Theorem. [Lardner's Edition]
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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;
For let AC be joined.
But the parallelogram ABCD is double of
the triangle ABC; Therefore, etc. Q.E.D. 
Proposition XLI. Theorem. [Lardner's Edition]
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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;
Let BC be bisected
at E, and let
AE be joined; Then FECG is a parallelogram.
And, since BE is equal to
EC,
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 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]
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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;
For since ABCD is a parallelogram,
and AC its diameter,
Again, since EH is a parallelogram,
and AK is its diameter,
For the same reason
Now, since the triangle AEK
is equal to the
triangle AHK,
And the whole triangle ABC is also
equal to the whole
ADC; Therefore etc. Q.E.D. 
Proposition XLIII. Theorem. [Lardner's Edition]
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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; 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,
Let them be produced and meet at K;
through the point K let
KL be drawn parallel to either
EA or FH,
[I. 31]
Then HLKF is a
parallelogram,
But BF is equal to the
triangle C;
And, since the angle GBE
is equal to the angle ABM,
[I. 15]
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]
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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;
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;
Then, since the angle E is equal
to each of the angles
HKF,
GHM,
Let the angle KHG be added
to each;
But the angles FKH,
KHG are equal to two right angles;
[I. 29]
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; 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;
But the angles
MHG, HGL
are equal to two right angles;
[I. 29]
And, since FK is equal and
parallel to HG,
[I. 34]
And, since the triangle ABD is equal
to the parallelogram
FH, 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]
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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;
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;
Therefore ADEB is a
parallelogram; I say next that it is also rightangled.
For, since the straight line AD
falls upon the parallels AB,
DE, 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]
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Euclid, Elements of Geometry, Book I, Proposition 47 

Proposition 47 [Heath's Edition]In rightangled 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 rightangled triangle
having the angle
BAC right;
For let there be described on BC
the square BDEC,
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;
For the same reason
And, since the angle DBC
is equal to the angle FBA: for
each is right:
And, since DB is equal to
BC, and
FB to
BA,
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]
Similarly, if
AE, BK
be joined, 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]
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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; 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,
Let the square on AD be
added to each;
But the square on DC
is equal to the squares on DA,
AC,
for the angle DAC is right;
[I. 47]
And since DA is equal to
AB, Therefore etc. Q.E.D. 
Proposition XLVIII. Theorem. [Lardner's Edition]
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^{1}(Note (hyp.) on the Proof of Proposition VII.) The hypothesis means the supposition; that is, the part of the enunciation of the proposition in which something is supposed to be granted true, and from which the proposed conclusion is to be inferred. Thus in the seventh proposition the hypothesis is, that the triangles stand on the same side of their base, and that their conterminous sides are equal, and the conclusion is a manifest absurdity, which proves that the hypothesis must be false.
In the fourth proposition the hypothesis is, that two sides and the included angle of one triangle are respectively equal to two sides and the included angle of the other; and the conclusion deduced from this hypothesis is, that the remaining side and angles in the one triangle are respectively equal to the remaining side and angles in the other triangle.
(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.