Given: ABC is a triangle and P is a point on the side BC such that AB = AP. AP produced meets the circumcircle of ∆ABC at Q.
To Prove: CP = CQ
Proof :
[ Angles in the same segment of a circle are equal ]
[ Vertically opposite angles ]
[ AA criterion of similarily ]
[ Corresponding sides of two similar triangles are proportional ]
But AB = AP | Given
CQ = CP
[ Angles opposite to equal sides of a triangle are equal ]