Question

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

Answer 1

Given: From a point $P$, two tangents $PA$ and $PB$ are drawn to a circle with center $O$. and $OP=$ diameter of the circle.

$\angle OAP=90°$ ($PA$ and $PB$ are the tangents to the circle.)

In $\Delta OPA$,
$sin\angle OPA=$ $\frac{OA}{OP}=\frac{r}{2r}$ [$OP$ is the diameter $=2\times radius$]
$sin\angle OPA=$ $\frac{1}{2}$ $=sin30⁰$
$\angle OPA=30°$

Similarly, $\angle OPB=30°$.
$\angle APB=\angle OPA+\angle OPB=30°+30°=60°$

In $\Delta PAB$,
$PA=PB$ (tangents from an external point to the circle)
$\Rightarrow \angle PAB=\angle PBA$ ............(1) (angles opp.to equal sides are equal)
$\Rightarrow \angle PAB+\angle PBA+\angle APB=180°$ [Angle sum property]
$\Rightarrow \angle PAB+\angle PAB=180°–60°=120°$ [Using (1)]
$\Rightarrow 2\angle PAB=120°$
$\Rightarrow \angle PAB=60°$ .............(2)

From (1) and (2)

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

All angles are equal in an equilateral triangle. ($60°$)

Hence, $\Delta PAB$ is an equilateral triangle.