Question

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

Answer 1

Let the two congruent circles with centre O and O' intersect each other at A and B.
Then, AB is common chord for both the circles.
$\angle AOB=\angle A{O}^{\prime }B........\left(1\right)$
Let P and Q be any two points in circles such that PAQ is a line segment.
It is known that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
$\angle AQB=\frac{1}{2}\angle AOB$
and
$\angle APB=\frac{1}{2}\angle A{O}^{\prime }B.........\left(2\right)$
From (1) and (2),
$\angle AQB=\angle APB$
It is given that PAQ is a line segment.
Therefore, $△PQB$ is an isosceles triangle where $\angle AQB=\angle APB$
Thus, $BQ=BP$