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

Refer pic.

Join AC,PQ and BD.

ACQP is a cyclic quadrilaterl.

$\angle CAP+\angle PQC={180}^{\circ }$....(1) pair of opposites in cyclic quadrilateral.

PQDB is a cyclic quadrilaterl.

$\angle PQD+\angle DBP={180}^{\circ }$....(2) pair of opposites in cyclic quadrilateral.

$\angle PQC+\angle PQD={180}^{\circ }$.....(3) CQD is a straight line.

From (1), (2) and (3), we get,

$\angle CAP+\angle DBP={180}^{\circ }$

$\angle CAB+\angle DBA={180}^{\circ }$

If a traversal intersects 2 lines such that a pair of interior angles on same side of traversal is supplimentary, then the 2 lines are parallel.

$\therefore AC\parallel BD$