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Question
Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at C and D. Prove that AC is parallel to BD.
Solution
Join AC, PQ and BD
ACQP is a cyclic quadrilateral
∴ ∠CAP + ∠PQC = 180° ...(i)
(Pair of opposite in a cyclic quadrilateral are supplementary)
PQDB is a cyclic quadrilateral
∴ ∠PQD + ∠DBP = 180° ...(ii)
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
Again, ∠PQC + ∠PQD = 180° ...(iii)
(CQD is a straight line)
Using (i), (ii) and (iii)
∴ ∠CAP + ∠DBP = 180°
Or ∴ ∠CAB + ∠DBA = 180°
We know, if a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel
∴ AC || BD
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