#### 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°

#### Solution

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°

Is there an error in this question or solution?

Solution 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° Concept: Tangent Properties - If Two Circles Touch, the Point of Contact Lies on the Straight Line Joining Their Centers.