AP and AQ are two tangents drawn from an external point A to a circle with centre O, P and Q are points of contact. If PR is a diameter, prove that `OA||RQ.`

Explanation

AP and AQ are two tangents to the circle with centre O and OP and OQ are two radii or the circle passing through poits of contact P and Q respectively.
`therefore OP bot AP and OQ bot AQ`
`therefore angle OPA =angle OQA=90^(@)`
Now, `Delta AOP -=Delta AOQ, ` [since `angle APO =angle AQO, OP =OQ and ` hypotenuse OA is common to both.]
`therefore angle AOP =angle AOQ [because` similar angles of congruent triangles] ......(1)

`therefore angle POQ =angle AOP+angle AOQ`
`= angle AOPQ+ angle AOQ ` [from (1)]
`=2 angle AOQ......(2)`
Again, `angle POQ` is the central angle and `angle PRQ` is the angle in circle produced by the arc PQ,
`therefore angle POQ =2 angle PRQ` [by theorem-34]
`or, 2 angle AOQ=2 angle PRQ `[from (2)]
`or, angle AOQ=angle PRQ=angle OQR [because OQ =OR, therefore angle ORQ=angle OQR]`
`thereforeangle AOQ =angle OQR.`
But these are two alternate angles when OQ intersects the line segments OA and RQ, and they are also equal.
`therefore OA ||RQ.` (Proved)