Pons asinorum

Euclid and Proclus

Euclid's statement of the pons asinorum includes a second conclusion that if the equal sides of the triangle are extended below the base, then the angles between the extensions and the base are also equal. Euclid's proof involves drawing auxiliary lines to these extensions. But, as Euclid's commentator Proclus points out, Euclid never uses the second conclusion and his proof can be simplified somewhat by drawing the auxiliary lines to the sides of the triangle instead, the rest of the proof proceeding in more or less the same way.

There has been much speculation and debate as to why Euclid added the second conclusion to the theorem, given that it makes the proof more complicated. One plausible explanation, given by Proclus, is that the second conclusion can be used in possible objections to the proofs of later propositions where Euclid does not cover every case. The proof relies heavily on what is today called side-angle-side (SAS), the previous proposition in the Elements, which says that given two triangles for which two pairs of corresponding sides and their included angles are respectively congruent, then the triangles are congruent.

Proclus' variation of Euclid's proof proceeds as follows: Let ⁠${\displaystyle \triangle ABC}$⁠ be an isosceles triangle with congruent sides ⁠${\displaystyle AB\cong AC}$⁠. Pick an arbitrary point ⁠${\displaystyle D}$⁠ on side ⁠${\displaystyle AB}$⁠ and then construct point ⁠${\displaystyle E}$⁠ on ⁠${\displaystyle AC}$⁠ to make congruent segments ⁠${\displaystyle AD\cong AE}$⁠. Draw auxiliary line segments ⁠${\displaystyle BE}$⁠, ⁠${\displaystyle DC}$⁠, and ⁠${\displaystyle DE}$⁠. By side-angle-side, the triangles ⁠${\displaystyle \triangle BAE\cong \triangle CAD}$⁠. Therefore ⁠${\displaystyle \angle ABE\cong \angle ACD}$⁠, ⁠${\displaystyle \angle ADC\cong \angle AEB}$⁠, and ⁠${\displaystyle BE\cong CD}$⁠. By subtracting congruent line segments, ⁠${\displaystyle BD\cong CE}$⁠. This sets up another pair of congruent triangles, ⁠${\displaystyle \triangle DBE\cong \triangle ECD}$⁠, again by side-angle-side. Therefore ⁠${\displaystyle \angle BDE\cong \angle CED}$⁠ and ⁠${\displaystyle \angle BED\cong \angle CDE}$⁠. By subtracting congruent angles, ⁠${\displaystyle \angle BDC\cong \angle CEB}$⁠. Finally ⁠${\displaystyle \triangle BDC\cong \triangle CEB}$⁠ by a third application of side-angle-side. Therefore ⁠${\displaystyle \angle CBD\cong \angle BCE}$⁠, which was to be proved.

Pappus

Proclus gives a much shorter proof attributed to Pappus of Alexandria. This is not only simpler but it requires no additional construction at all. The method of proof is to apply side-angle-side to the triangle and its mirror image. More modern authors, in imitation of the method of proof given for the previous proposition have described this as picking up the triangle, turning it over and laying it down upon itself. This method is lampooned by Charles Dodgson in Euclid and his Modern Rivals, calling it an "Irish bull" because it apparently requires the triangle to be in two places at once.

The proof is as follows: Let ABC be an isosceles triangle with AB and AC being the equal sides. Consider the triangles ABC and ACB, where ACB is considered a second triangle with vertices A, C and B corresponding respectively to A, B and C in the original triangle. ${\displaystyle \angle A}$ is equal to itself, AB = AC and AC = AB, so by side-angle-side, triangles ABC and ACB are congruent. In particular, ${\displaystyle \angle B=\angle C}$.

Others

A standard textbook method is to construct the bisector of the angle at A. This is simpler than Euclid's proof, but Euclid does not present the construction of an angle bisector until proposition 9. So the order of presentation of Euclid's propositions would have to be changed to avoid the possibility of circular reasoning.

The proof proceeds as follows: As before, let the triangle be ABC with AB = AC. Construct the angle bisector of ${\displaystyle \angle BAC}$ and extend it to meet BC at X. AB = AC and AX is equal to itself. Furthermore, ${\displaystyle \angle BAX=\angle CAX}$, so, applying side-angle-side, triangle BAX and triangle CAX are congruent. It follows that the angles at B and C are equal.

Legendre uses a similar construction in Éléments de géométrie, but taking X to be the midpoint of BC. The proof is similar but side-side-side must be used instead of side-angle-side, and side-side-side is not given by Euclid until later in the Elements.