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From a point $P$, two tangents $PA$ and $PB$ are drawn to a circle with centre $O$. If $OP$ is equal to the diameter of the circle, show that $\Delta APB $ is equiltateral.

So this is the figure:enter image description here

I have done the follwoing but am far away from the proof:

Let the radius = r cm

Then $OA = r \ cm$ and $OP = 2r\ cm$

Since $\angle OAP = 90 ^{\circ}$, therefore $AP = \sqrt3 r \ cm $

But I'm confused after that. What should I do?

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    $\begingroup$ Now get angle AOP, then get AB (or, half AB). $\endgroup$ Feb 23 '14 at 8:18
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given $OP=d$ (d for diameter) and $OA=\frac{d}{2}$. since $POA$ is a right triangle. $$\cos (\angle AOP)=\frac{OA}{OP}$$ you get $\angle AOP=60°$. now find all other angles with results you know about the figure. and finally prove $$\angle PAB= \angle APB=\angle PBA=60°$$

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