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.
┬а
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 тИЖOAP,
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)
тИЖOAQ is equilateral triangle тИаAOQ = 60┬░
Now in right тИЖOAP,
тИаAPO = 90┬░ тАУ 60┬░ = 30┬░
=> тИаAPB = 2 тИаAPO = 2 x 30┬░ = 60┬░
But PA = PB (Tangents from P to the circle)
=> тИаPAB = тИаPBA = 60┬░
Hence тИЖAPB is an equilateral triangle.