$AB$ is a line segment. $P$ and $Q$ are points on opposite sides of $AB$ such that each of them is equidistant from the points $A$ and $B$ (see figure). Show that the line $PQ$ is the perpendicular bisector of $AB$.

(N/A) Given: $PA = PB$ and $QA = QB$. We need to show that $PQ \perp AB$ and $PQ$ bisects $AB$. Let $PQ$ intersect $AB$ at $C$.
Consider $\Delta PAQ$ and $\Delta PBQ$:
$AP = BP$ (Given)
$AQ = BQ$ (Given)
$PQ = PQ$ (Common side)
Therefore,$\Delta PAQ \cong \Delta PBQ$ by $SSS$ congruence rule.
This implies $\angle APQ = \angle BPQ$ by $CPCT$.
Now,consider $\Delta PAC$ and $\Delta PBC$:
$AP = BP$ (Given)
$\angle APC = \angle BPC$ (Since $\angle APQ = \angle BPQ$)
$PC = PC$ (Common side)
Therefore,$\Delta PAC \cong \Delta PBC$ by $SAS$ congruence rule.
This implies $AC = BC$ and $\angle ACP = \angle BCP$ by $CPCT$.
Since $\angle ACP + \angle BCP = 180^o$ (Linear pair),
$2 \angle ACP = 180^o \implies \angle ACP = 90^o$.
Since $AC = BC$ and $\angle ACP = 90^o$,$PQ$ is the perpendicular bisector of $AB$.