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Question

In fig., OP is equal to diameter of the circle. Prove that ΔABP is an equilateral triangle.

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Solution

Join AB

Sin APO = OAOP = r2r

Sin APO = 12

APO = 30

Similarly

BPO = 30

APB = APO + BPO = 30 + 30 = 60

As the lengths of tangents drawn from an external point to a circle are equal,

PA = PB

PAB = PBA

In ΔPAB, ABP + BAP + APB = 180

2ABP + 60 = 180

ABP =60

Therefore BAP = 60

ΔPAB is an equilateral triangle.


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