Question

Two circles intersect each other at the points A and B ; and one circle passes through the centre of the other. P is any point on the first circle and the straight line PQ cuts the other circle at Q. Prove that PB = PQ

Please Help me with this question!! ​

Answer 1

Step-by-step explanation:

Join AC,PQ and BD.

ACQP is a cyclic quadrilaterl.

∠CAP+∠PQC=180

∘

....(1) pair of opposites in cyclic quadrilateral.

PQDB is a cyclic quadrilaterl.

∠PQD+∠DBP=180

∘

....(2) pair of opposites in cyclic quadrilateral.

∠PQC+∠PQD=180

∘

.....(3) CQD is a straight line.

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

∠CAP+∠DBP=180

∘

∠CAB+∠DBA=180

∘

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.

∴AC∥BD