Two circles intersect each other at the points P and Q. Two straight lines through P and Q intersect on circle at the points A and C and the other circle at B and D . Prove that AC||BD.

Explanation

we know that two opposite angles of a cyclic quadrilateral are supplementary.
`therefore` for the quadrilateral ACQP,
`angleCAP+anglePQC=180^(@)" ……..(1) "`
and for the quadrilateral BPQD
`anglePBD+anglePQD=180^(@)` …….(2)
Now adding (1) and(2) , we get ` angleCAP+anglePBD+anglePQC+anglePQD=360^(@)`
or, `angleCAP+anglePBD+angleCQP=360^(@)[becauseanglePQC+anglePQD=angleCQD=1 "straight angle "=180^(@)`
or , `angleCAP+anglePBD+180^(@)=360^(@)"or",angleCAP+anglePBD=360^(@)-180^(@)`
`angleCAP+anglePBD=180^(@)`
But the common transversal of the straight lines AC and BD is AB and two adjacent angles on the same side of AB are `angleCABand angleDBA` , the sum of which is `180^(@)`
AC||BD[`because` the sum of two adjacent angles on the same side of the common transversal of two straight lines `180^(@)` ]
Hence AC||BD .