Angle bisector theorem

Theorem 2

Prove that: If a point is equidistant from the endpoints of a line segment, then it lies on the perpendicular bisector of the segment. (Converse of theorem 1)

Given: $CD$ is the perpendicular bisector of $\overline{AB}$ intersecting it at $O$. $P$ is any point on $\overline{AB}$ such that $\overline{AP} \cong \overline{BP}$.

To Prove: $P$ lies on the perpendicular bisector of $\overline{AB}$, i.e. $OP \perp AB$.

Construction: Bisect $\overline{AB}$ at $O$. Join points $P$ and $O$.