If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.

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 unqueal 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.