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

Ya then it's very easy
but wait for his response
The same sum is given in my textbook and it's written that circles intersect at A and B and P and Q lie on the two circles.
I will write the answer accordingly


See the pic for reference.

Chord AB will subtend equal angles on both the circles' circumferences.
So, Angle APB = Angle AQB
So, in triangle BPQ,
QB = PB (Because they are sides opposite to equal angles)
Hence proven.
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Please mark this as the best answer
AB is a common cord  of congruent circles
.'. angle subtended by that chord are equal
.'. angle AQB = angle APB

now in ΔPQB
'.' angle AQB = angle APB

.'.  BQ = BP        ( because sides opposite to equal angles are equal)
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