Two circles intersect each other at the points P and Q. A straight line passing through P intersects one circle at A the other circle at .A straight line passing through Q intersect the first circle at C and the second circle at D. Prove that `AC||BD`.

Explanation

Let two circles with centres at O adn O' intersect each other at the points P and Q respectively. A straight ine pasing through P intersect the circle with centre O at the point A and the circle with centre O' at the point B. Another straight line pasising through Q also intersects the first circle at C and the secondary circle at D.
We have to prove that `AC||BD`.
Construction : Let us join P,Q
Proof : APQC is a cyclic quadrilateral.
`:. angle PAC+anglePQC=180^(@)....(1) [ :.` opposite angles of cyclic quadrilateral are supplementary.]
Again BPQD is also a cyclic quadrilaterla.
`:. angle PBD+angle PQD=180^(@)......(2)` [ for smililar reason]
Now, addinng (1) and (2) we get, ,
`anglePAC+anglePBD+ anglePQC +anglePQD=360^(@)`
` or, angle PAC+ angle PBD+180^(@)=360^(@) [ :. angle PQC+ anglePQD=1` straight angle `=180^(@)`]
or `angle PAC+ angle PBD=180^(@)`
i.e. the sum of two adjacent angles on the same side of the side AB of the transvrsal AB of the two line segements AC and BD is `180^(@)`.
Hence `AC||BD`