Asked by kirtichawla94 | 20th Nov, 2009, 07:08: PM
Let OP meets the circle at a point Q.
It is given that OP = 2r
So, OP OP = QP + OP
QP = r.
Therfore, Q is the mid point of OP.
Now AP is tnagent at A and OA is radius
OA is perpendicular to AP.
Therefore, Triangle OAP is right triangle at A . Q is the mid point of hypotnuese OP.
Therfore, AQ = OQ = OA
i.e Triangle OAQ is equilateral.
So, AOQ = 60o
Therefore, APO = 180o - 90o- 60o = 30o
APB = 2APO = 60o
Also, PA = PB
PAB = PBA
Now in triangle APB, we have
PAB + PBA + APB = 180o
PAB = 60o
So, triangle APB is an equilateral triangle.
Answered by | 24th Nov, 2009, 10:36: AM
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