Given: From a point P outside the circle with centre O, PA and PB are the tangents to the circle such that OP is diameter.
And, AB is joined.
Required to prove: APB is an equilateral triangle
Construction: Join OP, AQ, OA
Proof:
We know that, OP = 2r
⇒ OQ + QP = 2r
⇒ OQ = QP = r
Now in right ∆OAP,
OP is its hypotenuse and Q is its mid-point
Then, OA = AQ = OQ
(mid-point of hypotenuse of a right triangle is equidistances from its vertices)
Thus, ∆OAQ is equilateral triangle. So, ∠AOQ = 60°
Now in right ∆OAP,
∠APO = 90° – 60° = 30°
⇒ ∠APB = 2 ∠APO = 2 x 30° = 60°
But PA = PB (Tangents from P to the circle)
⇒ ∠PAB = ∠PBA = 60°
Hence ∆APB is an equilateral triangle.