Two congruent circles intersect each other at points $A$ and $B$. Through $A$,any line segment $PAQ$ is drawn such that $P$ and $Q$ lie on the two circles. Prove that $BP = BQ$.

(N/A) Given: Two congruent circles intersect at $A$ and $B$. $A$ line segment $PAQ$ passes through $A$,where $P$ lies on the first circle and $Q$ lies on the second circle.
Construction: Join $AB$,$BP$,and $BQ$.
Proof:
$1$. Consider the two congruent circles. The chord $AB$ is common to both circles.
$2$. In congruent circles,equal chords subtend equal angles at the circumference.
$3$. Since $AB$ is a chord in both congruent circles,the angles subtended by $AB$ at the circumference in the respective circles must be equal.
$4$. Therefore,$\angle APB = \angle AQB$.
$5$. In $\Delta PBQ$,we have $\angle APB = \angle AQB$ (which is $\angle BPQ = \angle BQP$).
$6$. Since the angles opposite to sides $BQ$ and $BP$ are equal,the sides themselves must be equal.
$7$. Hence,$BP = BQ$.