# 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 $\mathrm{BP}=\mathrm{BQ}$.

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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.

Updated on 10-Oct-2022 13:46:58