In isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another.

Let *ABC* be an isosceles triangle having the side *AB* equal to the side *AC,* and let the straight lines *BD* and *CE* be produced further in a straight line with *AB* and *AC.*

I say that the angle *ABC* equals the angle *ACB,* and the angle *CBD* equals the angle *BCE.*

Take an arbitrary point *F* on *BD.* Cut off *AG* from *AE* the greater equal to *AF* the less, and join the straight lines *FC* and *GB.*

Since *AF* equals AG, and *AB* equals *AC,* therefore the two sides *FA* and *AC* equal the two sides *GA* and * AB,* respectively, and they contain a common angle, the angle *FAG.*

Therefore the base *FC* equals the base *GB,* the triangle *AFC* equals the triangle *AGB,* and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides, that is, the angle *ACF* equals the angle *ABG,* and the angle *AFC* equals the angle *AGB.*

Since the whole *AF* equals the whole *AG,* and in these *AB* equals *AC,* therefore the remainder *BF* equals the remainder *CG.*

But *FC* was also proved equal to *GB,* therefore the two sides *BF* and *FC* equal the two sides *CG* and *GB* respectively, and the angle *BFC* equals the angle *CGB,* while the base *BC* is common to them. Therefore the triangle *BFC* also equals the triangle *CGB,* and the remaining angles equal the remaining angles respectively, namely those opposite the equal sides. Therefore the angle *FBC* equals the angle *GCB,* and the angle *BCF* equals the angle *CBG.*

Accordingly, since the whole angle *ABG* was proved equal to the angle *ACF,* and in these the angle *CBG* equals the angle *BCF,* the remaining angle *ABC* equals the remaining angle *ACB,* and they are at the base of the triangle *ABC.* But the angle *FBC* was also proved equal to the angle *GCB,* and they are under the base.

Therefore *in isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another.*

Q.E.D.

Unfortunately, such an argument would be circular. I.13 depends on I.11, I.11 on I.8, I.8 on I.7, and I.7 on I.5. Thus, I.13 cannot be used in the proof of I.5. It may appear that I.7 only depends on the first conclusion of I.5, but a case of I.7 that Euclid does not discuss relies on the second conclusion of I.5.

This proposition has been called the *Pons Asinorum*, or Asses’ Bridge. Whether this name is due to its difficulty (which it isn’t) or the resemblance of its figure to a bridge is not clear. Very few of the propositions in the *Elements* are known by names.

The difficulty lies in treating one triangle as two, or in making a correspondence between a triangle and itself, but not the correspondence of identity. There is nothing wrong with this proof formally, but it might be more difficult for a student just learning geometry.