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.

Answer 1

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