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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