In figure (i) given below, P is the point of intersection of the chords BC and AQ such that AB = AP. Prove that CP = CQ.

Given, two chords AQ and BC intersect each other at P inside the circle. AB and CQ are joined and AB = AP.

To prove : CP = CQ

Construction : Join AC.

Proof :

In △ABP and △CQP

∠B = ∠Q (∵ angles in same segment are equal)

∠BAP = ∠PCQ (∵ angles in same segment are equal)

∠BPA = ∠CPQ (∵ vertically opposite angles are equal.)

∴ △ABP ~ △CQP (By AAA axiom of similarity.)

Since, triangles are similar hence, the ratio of the corresponding sides are equal.

$\dfrac{AB}{CQ} = \dfrac{AP}{CP}$

We know, AB = AP,

$\therefore \dfrac{AP}{CQ} = \dfrac{AP}{CP} \\[1em] CQ = CP.$

Hence, proved that CQ = CP.