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.

Here is the proof to the mentioned result.

Let the two congruent (of equal radii) circles with centre O and O' intersect each other at A and B.

Then, AB is common cord for both the circles.

∠AOB = ∠A'OB … (1)

Let P and Q be any two points on 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.

From (1) and (2),

∠AQB = ∠APB

It is given that PAQ is a line segment.

Therefore, ΔPQB is isosceles where ∠AQB = ∠APB.

Thus, BQ = BP.

Hence, the result is proved.

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