From a point `P`, two tangents `PA` and `PB` are drawn to a circle `C(O,r)`. If `OP=2r`, show that `DeltaAPB` is equilateral.

Explanation

Let `OP` meet the circle at `Q`.
Join `OA` and `AQ`.
Clearly, `OA bot APimplies/_OAP=90^(@)`[radius through the point of contact is perpendicular to the tangent].
Now, `OQ=QP=r`
Thus, `Q` is the midpoint of the hypotenuse `OP` of `DeltaOAP`.
So, `Q` is equidistant from `O`, `A` and `P`.
`:.QA=OQ=QP=r`
`impliesOA=OQ=QA=r`
`impliesDeltaAOQ` is equilateral
`implies/_AOQ=60^(@)` [`:'` each angle of an equilateral triangle is `60^(@)`]
`implies/_AOP=60^(@)`
`implies/_APO=30^(@)` [`:' /_AOP+/_OAP+/_APO=180^(@)`]
`implies /_APB=2/_APO=60^(@)`.
Also, `PA=PBimplies/_PAB=/_PBA=60^(@)`
Hence, `DeltaPAB` is an equilateral triangle.