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 triangles 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 respectivel 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 ot 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. 

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. 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, 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 ot 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 EC D and FD 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 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 C G B and C B G 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 C G 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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 

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] 
^{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.