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.