Pons asinorum

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.[7] The proof relies heavily on what is today called side-angle-side, the previous proposition in the Elements.

Proclus' variation of Euclid's proof proceeds as follows:[8]

Let ABC be an isosceles triangle with AB and AC being the equal sides. Pick an arbitrary point D on side AB and construct E on AC so that AD = AE. Draw the lines BE, DC and DE.

Consider the triangles BAE and CAD; BA = CA, AE = AD, and ${\displaystyle \angle A}$ is equal to itself, so by side-angle-side, the triangles are congruent and corresponding sides and angles are equal.

Therefore ${\displaystyle \angle ABE=\angle ACD}$ and ${\displaystyle \angle ADC=\angle AEB}$, and BE = CD.

Since AB = AC and AD = AE, BD = CE by subtraction of equal parts.

Now consider the triangles DBE and ECD; BD = CE, BE = CD, and ${\displaystyle \angle DBE=\angle ECD}$ have just been shown, so applying side-angle-side again, the triangles are congruent.

Therefore ${\displaystyle \angle BDE=\angle CED}$ and ${\displaystyle \angle BED=\angle CDE}$.

Since ${\displaystyle \angle BDE=\angle CED}$ and ${\displaystyle \angle CDE=\angle BED}$, ${\displaystyle \angle BDC=\angle CEB}$ by subtraction of equal parts.

Consider a third pair of triangles, BDC and CEB; DB = EC, DC = EB, and ${\displaystyle \angle BDC=\angle CEB}$, so applying side-angle-side a third time, the triangles are congruent.

In particular, angle CBD = BCE, which was to be proved.

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.[9][6] This method is lampooned by Charles Lutwidge 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.[10]

The proof is as follows:[11]

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, ${\displaystyle \angle B=\angle C}$.[12]

A textbook proof

A standard textbook method is to construct the bisector of the angle at A.[13] 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 the Euclid's propositions would have to be changed to avoid the possibility of circular reasoning.

The proof proceeds as follows:[14]

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.[15] 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.

In 1876, while a member of the United States Congress, future President James A. Garfield developed a proof using the trapezoid, which was published in the New England Journal of Education.[16] Mathematics historian William Dunham wrote that Garfield's trapezoid work was "really a very clever proof."[17] According to the Journal, Garfield arrived at the proof "in mathematical amusements and discussions with other members of congress."[18]