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 = 180°

Answer 1

Draw Seg PQ.
APQD is a cyclic qudrilateral.
∠ ADQ + ∠ APQ = 180° ....... (1)
PBCQ is a cyclic qudrilateral.
∴ ∠ BCQ + ∠ BPQ = 180° ...... (2)
∴ ∠ ADQ + ∠ APQ + ∠ BCQ + ∠ BPQ = 180° +180°.... [from (1),(2) ]
∴ ∠ ADQ + ∠ BCQ + ∠ APQ + ∠ BPQ = 180°+ 180° ...... (3)
But∠ APQ + ∠ BPQ = 180° ............ (4) (angles in linear pair)

∴ ∠ ADQ + ∠ BCQ + 180° = 180° + 180° ............ [from (3) , (4) ]
∴ ∠ ADQ + ∠ BCQ = 180°
∴ ∠ ADC + ∠ BCD = 180°