Question

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

Answer 1

REF.Image

AP is the tangent to the circle.

$\therefore OA\perp AP$ (Radius is perpendicular to the tangent at the point of contact)

$\Rightarrow \angle OAP={90}^{\circ }$

In $△OAP$,

$sin\angle OPA=\frac{OA}{OP}=\frac{r}{2r}$ [OP = diameter]

$\therefore sin\angle OPA=\frac{1}{2}={30}^{\circ }$

$\Rightarrow \angle OPA={30}^{\circ }$

Similarly, it can be proved that $\angle OPB={30}^{\circ }$

Now, $\angle APB=\angle OPA+\angle OPB={30}^{\circ }+{30}^{\circ }={60}^{\circ }$

IN $△PAB$,

$PA=PB$ [length of tangents drawn from external point]

$\Rightarrow \angle PAB=\angle PBA......\left(i\right)$ [equal sides have equal angles opp. to them]

$\angle PAB+\angle PBA+\angle APB={180}^{\circ }$ [angle sum property]

$\Rightarrow \angle PAB+\angle PBA={180}^{\circ }-{60}^{\circ }$

from (i)

$\Rightarrow 2\angle PAB={120}^{\circ }$

$\Rightarrow \angle PAB={60}^{\circ }......\left(ii\right)$

from (i) and (ii)

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

Therefore, PAB is an equilateral triangle.