Two circles have been drawn with centres A and B which touch each other externally at the point O. A straight line is drawn passing thrugh the point O and intersects the two circles at P and Q respectively. Prove that `AP ||BQ.`

Explanation

Let two circles with circles with centres A and B touch each other externally at the point O.
`therefore A, O, B` will lie on the same straight line.
Now, in `DeltaAOP and Delta BOQ, angle AOP =angle BOQ [because ` Oppostie angles]
Again, in circle wit centre at A, AO =AP `[because` radii of same circle].........(1)
Also, in circle with centre at B,
`BO =BQ[because` Radii of same circle]
`therefore angle BOQ=angleBQO..........(2)`
Since `angle AOP=angle BOQ, ` from (1) and (2) we get, `angle APO=angleBQO`
But these are the alternate angles produced when two line segments AP and BQ are intersected by the transversal PQ and also they are equal.
`therefore AP ||BQ` (Proved)