Given: O is the centre From a point P outside the circle, PA and PB are the tangents to the circle such that OP is diameter of the circle.

And, AB is joined.

Given to prove: APB=equilateral triangle

Construction: Join OP, AQ, OA

Proof:

We know that, $OP=2r$

$\Rightarrow OQ+QP=2r$

$\Rightarrow OQ=QP=r$

Now in right$\vartriangle OAP$.

$OP$  is the hypotenuse and $Q$ is mid-point of $\vartriangle OAP$

Then, $OA=AQ=OQ$

(In a right angled triangle mid-point of hypotenuse is equidistance from its vertices)

Thus, $\vartriangle OAQ$ is equilateral triangle. So, $\angle AOQ={{60}^{\circ }}$

Now in right$\vartriangle OAP$,

$\angle APO={{90}^{\circ }}-{{60}^{\circ }}={{30}^{\circ }}$

$\Rightarrow \angle APB=2\angle APO$

$\Rightarrow \angle APB=2\times {{30}^{\circ }}={{60}^{\circ }}$

But $PA=PB$

$\Rightarrow \angle PAB=\angle PBA={{60}^{\circ }}$

Hence $\vartriangle APB$  is an equilateral triangle.