Question

Two circles intersect each other at points

P and Q. Secants drawn through P and Q

intersect the circles at points A,B and

D,C.

Prove that : ∠ADC + ∠BCD = 180o

Answer 1

Answer:

It would appear that the points A, B, C and D are such that A and D are on one circle, while B and C are on the other.  Let's proceed from here.

Since PBCQ is then a cyclic quadrilateral, and opposite angles in a cyclic quadrilateral are supplementary, we have

∠BCD = ∠BCQ = 180° - ∠BPQ = ∠APQ.

Since APQD is cyclic, we likewise have

∠APQ = 180° - ∠ADQ = 180° - ∠ADC.

Hence ∠ADC + ∠BCD = 180°.

Answer 2

hence proved

hope this helps you