From a point P, two tangents …

Join OP, Suppose OP meets the circle at Q. Join AQ.

We have,
OP = diameter
{tex} \Rightarrow{/tex} OQ + PQ = diameter
{tex} \Rightarrow{/tex} PQ = diameter - radius
{tex} \Rightarrow{/tex} PQ = radius.
Therefore, OQ = PQ = radius.
Thus, OP is the hypotenuse of right triangle OAP and Q is the mid-point of OP.
Now,
In {tex} \triangle{/tex}OPA,
{tex} \sin \angle OPA = \frac{{OA}}{{OP}} = \frac{r}{{2r}}{/tex} [Give OP is the diameter of the circle]
{tex} \Rightarrow \sin \angle OPA = \frac{1}{2} = \sin 30^\circ{/tex}
{tex} \Rightarrow \angle OPA = 30^\circ{/tex}
Similarly, it can be proved that {tex} \angle OPB = 30^\circ {/tex}.
Now, {tex} \angle APB = \angle OPA + \angle OPB{/tex} {tex}= 30^\circ + 30^\circ = 60^\circ{/tex}
In {tex} \triangle{/tex}PAB,
PA = PB [Lengths of tangents drawn from an external point to a circle are equal]
{tex}\Rightarrow \angle PAB = \angle PBA{/tex} ....(i) [Equal sides have equal angles opposite to them]
{tex} \angle PAB + \angle PBA + \angle APB = 180^\circ {/tex} [Angle sum property]
{tex}\Rightarrow \angle PAB + \angle PAB = 180^\circ - 60^\circ = 120^\circ {/tex} [Using (i)]
{tex} \Rightarrow 2\angle PAB = 120^\circ{/tex}
{tex} \Rightarrow \angle PAB = 60^\circ{/tex} ....(ii)
From (i) and (ii)
{tex} \angle PAB = \angle PBA = \angle APB = 60^\circ {/tex}
{tex}{/tex} Therefore, {tex} \triangle{/tex}PAB is an equilateral triangle.