#### Question

Two circle 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