Question.
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.
Solution:
$\mathrm{AB}$ is the common chord in both the congruent circles.
$\therefore \angle \mathrm{APB}=\angle \mathrm{AQB}$
In $\triangle B P Q$,
$\angle \mathrm{APB}=\angle \mathrm{AQB}$
$\therefore B Q=B P$ (Angles opposite to equal sides of a triangle)
$\mathrm{AB}$ is the common chord in both the congruent circles.
$\therefore \angle \mathrm{APB}=\angle \mathrm{AQB}$
In $\triangle B P Q$,
$\angle \mathrm{APB}=\angle \mathrm{AQB}$
$\therefore B Q=B P$ (Angles opposite to equal sides of a triangle)
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