Point $M$ lies on the perpendicular bisector of $PQ$. Also $M$ does not lie on $PQ$. Prove that $M$ is equidistant from $P$ and $Q$.

(N/A) Let $PQ$ be a line segment and $L$ be its perpendicular bisector. Let $M$ be a point on $L$ such that $M$ does not lie on $PQ$. Let $L$ intersect $PQ$ at point $O$. Since $L$ is the perpendicular bisector of $PQ$,we have $PO = OQ$ and $\angle MOP = \angle MOQ = 90^{\circ}$. Consider $\triangle MOP$ and $\triangle MOQ$. In these triangles: $1$. $PO = OQ$ (Given). $2$. $\angle MOP = \angle MOQ = 90^{\circ}$ (Given). $3$. $MO = MO$ (Common side). By the $SAS$ (Side-Angle-Side) congruence criterion,$\triangle MOP \cong \triangle MOQ$. Since the triangles are congruent,their corresponding parts are equal $(CPCT)$. Therefore,$MP = MQ$. Thus,point $M$ is equidistant from $P$ and $Q$.