The Elements

...the angle d.b.c. will be made equal to angle e.c.b., which is proven as follows: having extended a.b. and a.c., I will set—by the third [proposition]—line a.d. equal to line a.e., and I will draw lines e.b. and d.c. And I understand two triangles, a.b.e. and a.c.d., which I will prove to be equal and equilateral original: "equilateros"; in this context, it means the corresponding sides of the two triangles are equal, not that the triangles themselves have three equal sides and equiangular. For the two sides a.b. and a.e. of triangle a.b.e. are equal to the two sides a.c. and a.d. of triangle a.c.d., and angle a is common to both; therefore, by the preceding [proposition], the base b.e. is equal to base c.d., and angle e is equal to angle d, and angle a.b.e. is equal to angle a.c.d. Likewise, I understand two triangles d.b.c. and e.c.b., which I will similarly prove to be equilateral and equiangular; for the two sides d.b. and d.c. of triangle b.d.c. are equal to the two sides e.c. and e.b. of triangle e.c.b., and angle d is equal to angle e. Therefore, by the preceding proposition: base to base, and the remaining angles to the remaining angles. Thus, angle d.b.c. is equal to angle e.c.b., and this is according to the proposition: namely, that the angles under the base are equal. And angle b.c.d. is equal to e.b.c., but the whole a.b.e. is equal to a.c.d., as was proven above. Therefore, the remaining angle a.b.c. is equal to the remaining angle a.c.b., which is also upon the base; which was the first part proposed.

A geometric diagram illustrating the proof of the equality of angles at the base of an isosceles triangle (Euclid's Elements, Book I, Prop. 5). It shows an isosceles triangle abc with its sides ab and ac extended. Auxiliary lines connect points d and e on the extensions to the opposite base vertices, forming two large overlapping triangles.

If two angles of any triangle are equal, then the two sides subtending original: "respicientia"; literally 'looking at' or 'opposite to' those angles will also be equal.
This is the converse of the preceding [proposition] as far as its first part is concerned. For let there be a triangle a.b.c. whose two angles b and c are equal. I say that side a.b. is equal to side a.c. For if they are not equal, one of them will be greater than the other. Let a.b. be the greater, which is cut off to the equality of a.c. by the third proposition, so that the excess is from the side of a, and let it be cut at point d; and let b.d. be equal to a.c., and I will draw line d.c. I understand, therefore, two triangles, a.b.c. and d.b.c., which I will prove to be equilateral and equiangular. For the two sides d.b. and b.c. of triangle d.b.c. are equal to the two sides a.c. and b.c. of triangle a.b.c., and angle b is equal to the whole angle c by hypothesis. Therefore, base d.c. is equal to base b.a., and angle d.c.b. is equal to angle a.c.b.—namely, a part to its whole—which is impossible This is a proof by contradiction: a part cannot be equal to the whole, therefore the initial assumption that the sides were unequal must be false..

A geometric diagram for Proposition 6 showing a triangle abc with an interior point d on side ab connected to vertex c, used in a proof by contradiction.

If from two points terminating a line, two lines are drawn meeting at a single point, it is impossible for two other lines, equal to the first two respectively and meeting at another point, to be drawn from the same points on the same side.
Let there be line a.b., from whose endpoints two lines are drawn on one side which meet at the same point, such as a.c. and b.c., which meet at point c. I say that on the same side, two other lines will not be drawn from the same endpoints meeting at another point, such that the one proceeding from point a is equal to line a.c. and the one proceeding from point b is equal to line b.c. For if it were possible, let two other lines be drawn on the same side meeting at point d, and let line a.d. be equal to line a.c. and line b.d. be equal to b.c. Therefore, point d will fall either inside the triangle or outside; for it will not fall on either of the sides a.c. or b.c., because then a part would be equal to its whole. But if it falls outside, either one of the lines a.d. and b.d. will cut one of the lines a.c. and b.c., or neither will cut the other. And let one first cut the other, and let line c.d. be drawn. Because the two sides a.c. and a.d. of triangle a.c.d. are equal, angle a.c.d. will be equal to angle a.d.c. by the fifth [proposition]. Similarly, because in triangle b.c.d. the two sides b.c. and b.d. are equal, the angles b.c.d. and b.d.c. will be [equal]...

A geometric diagram for Proposition 7 showing a base line ab with two distinct points c and d above it. Lines are drawn from the endpoints a and b to both c and d, and a connecting line cd is drawn between the two peaks.