Question -
Answer -
Given : From a point P outside the circle with centre O, PA and PB are the tangerts to the circle such that OP is diameter.
AB is joined.
To prove: APB is an equilateral triangle
Const : Join OP, AQ, OA
Proof : OP = 2r
=> OQ + QP = 2r
=> OQ = QP = r (OQ = r)
Now in right ∆OAP,
OP is its hypotenuse and Q is its mid point
OA = AQ = OQ
(mid-point of hypotenuse of a right triangle is equidistances from its vertices)
∆OAQ is equilateral triangle ∠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.