So, I'm trying to prove the statement in the title of this post. So far, I have developed the following proof, but I am unsure if it is correct.
Proof: Let $\overline{AB}$ be a line segment with perpendicular bisector $\overleftrightarrow{DE}$ intersecting $\overline{AB}$ at its midpoint $M$. Now, let $P$ be a point on $\overleftrightarrow{DE}$. If $P$ also lies on $\overline{AB}$, then $P = M$ since $M$ is the only point which lies on both $\overline{AB}$ and $\overleftrightarrow{DE}$. But $P=M$ is equidistant from $A$ and $B$ by definition. Therefore, we consider the case where $P$ is not on $\overline{AB}$. Since $P$ lies on $\overleftrightarrow{DE}$ but not on $\overline{AB}$, $A, M,$ and $P$ are non-collinear, so we may consider triangles $\triangle AMP$ and $\triangle BMP$. Trivially, we note that $\overline{MP} \cong \overline{MP}$, and since $M$ is the midpoint of $\overline{AB}$, we have $AM = BM$. It follows that $\overline{AM} \cong \overline{BM}$. Also, we note that the line segment $\overline{PM}$ lies on the line $\overleftrightarrow{DE}$, the perpendicular bisector of $\overline{AB}$. Therefore, $\overline{PM}$ must intersect $\overline{AB}$ at a right angle. Thus,
$m(\angle AMP) = m(\angle BMP) = 90 \implies \angle AMP \cong \angle BMP$.
Since $\angle AMP$ is the included angle of sides $\overline{AM}$ and $\overline{PM}$ of triangle $\triangle AMP$ and since $\angle BMP$ is the included angle of sides $\overline{BM}$ and $\overline{PM}$ of triangle $\triangle BMP$, we may conclude that $\triangle AMP \cong \triangle BMP$ by the SAS congruence condition. It follows that $\overline{PA} \cong \overline{PB} \implies PA = PB$, as desired. $\square$
Is everything okay with this proof or are there any missing details that perhaps I have failed to consider/account for?