Question

From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP = diameter of the circle, show that ∆APB is equilateral.

Answer 1

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.