Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see figure). Prove that \(\angle{ACP}\) = \(\angle{QCD}\).

Given:
Two circles intersect at two points B and C. Through B two line segment ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively.

To prove: \(\angle{ACP}\) = \(\angle{QCD}\)

Proof:
In circle I,
\(\angle{ACP}\) = \(\angle{ABP}\) ...(i)
(Angles in the same segment)

In circle II,
\(\angle{QCD}\) = \(\angle{QBD}\) ...(ii)
(Angles in the same segment)

Also, \(\angle{ABP}\) = \(\angle{QBD}\)
(Vertically opposite angles)

From Equation (i) and (ii), we get,

\(\angle{ACP}\) = \(\angle{QCD}\)
Hence, proved.