Q:

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?

A:

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

