If the perpendicular bisector of a chord $AB$ of a circle $PXAQBY$ intersects the circle at $P$ and $Q$,prove that arc $PXA \cong$ arc $PYB$.

(N/A) Given: $PQ$ is the perpendicular bisector of chord $AB$ of a circle,intersecting the circle at $P$ and $Q$. Let $M$ be the point of intersection of $PQ$ and $AB$.
To prove: Arc $PXA \cong$ Arc $PYB$.
Proof:
In $\Delta APM$ and $\Delta BPM$:
$1$. $AM = BM$ (Since $PQ$ is the perpendicular bisector of $AB$)
$2$. $\angle AMP = \angle BMP = 90^{\circ}$ (Given)
$3$. $PM = PM$ (Common side)
Therefore,$\Delta APM \cong \Delta BPM$ by the $SAS$ congruence rule.
This implies $AP = BP$ by $CPCT$.
Since $AP$ and $BP$ are chords of the circle and $AP = BP$,their corresponding arcs are congruent.
Thus,arc $PXA \cong$ arc $PYB$.