Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that `angle APB = 2 angle OAB`

Explanation

GIVEN A circle with centre `O` and `PA`, `PB` are the tangents on it from a point `P` outside it.
TO PROVE `/_APB=2/_OAB`.
PROOF Let `/_APB=x^(@)`
We know that the tangents to a circle from an external point are equal. So, `PA=PB`
Since the angles opposite to the equal sides of a triangle are equal, so
`PA=PBimplies/_PBA=/_PAB`.
Also, the sum of the angles of a triangle is `180^(@)`.
`:. /_APB+/_PAB+/_PBA=180^(@)`
`impliesx^(@)+2/_PAB=180^(@)` [ `:' /_PBA=/_PAB`]
`implies/_PAB=(1)/(2)(180^(@)-x^(@))=(90^(@)-(1)/(2)x^(@))`.
But , `PA` is a tangent and `OA` is the radius of the given circle.
`:. /_OAB+/_PAB=90^(@)`
`implies/_OAB=90^(@)-(90^(@)-(1)/(2)x^(@))implies/_OAB=(1)/(2)x^(@)=(1)/(2)/_APB`
`implies/_APB=2/_OAB`.