# Prove $\Delta APB$ is equilateral triangle

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:

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?

• Now get angle AOP, then get AB (or, half AB). Feb 23 '14 at 8:18

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 $$\angle PAB= \angle APB=\angle PBA=60°$$