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

Given:

Two congruent circles intersect each other at points \( \mathrm{A} \) and \( \mathrm{B} \). Through \( \mathrm{A} \) any line segment \( \mathrm{PAQ} \) is drawn so that \( \mathrm{P}, \mathrm{Q} \) lie on the two circles.

To do:

We have to prove that \( \mathrm{BP}=\mathrm{BQ} \).

Solution:

Let two circles intersect each other as shown in the below figure.

We know that,

Angles subtended by equal chords are equal.

Therefore,

$AB$ is the common chord in both the congruent circles

This implies,

$\angle APB = \angle AQB$

In $\triangle BPQ$,

$\angle APB = \angle AQB$

We know that,

Sides opposite to equal angles of a triangle are equal.

This implies,

$BQ = BP$

Hence proved.

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