Let ABC be a triangle having the angle ABC equal to the angle ACB;

I say that the side AB is also equal to the side AC.

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;

let DC be joined.

Then, since DB is equal to AC, and BC is common,

• the two sides DB, BC are equal to the two sides AC, CB respectively;

and the angle DBC is equal to the angle ACB;

• therefore the base DC is equal to the base AB, and the triangle DBC will be equal to the triangle ACB, the less to the greater: which is absurd. Therefore AB is not unequal to AC; it is therefore equal to it.

Therefore etc.

• Q. E. D.