Question

In figure, OP is equal to diameter of the circle. Prove that ABP is a equilateral triangle

Answer 1

PA and PB are the tangents to the circle.

∴ $OA\perp PA$

⇒ $\angle OAP=90°$

In $\Delta OPA$,

$sin\angle OPA=\frac{OA}{OP}=\frac{r}{2r}$ [Given OP is the diameter of the circle]

⇒ $sin\angle OPA=\frac{1}{2}=sin30⁰$

⇒ $\angle OPA=30°$

Similarly, it can be proved that $\angle OPB=30°.$

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

In ΔPAB,

$PA=PB$ [lengths of tangents drawn from an external point to a circle are equal]

⇒$\angle PAB=\angle PBA$ ............(1) [Equal sides have equal angles opposite to them]

$\angle PAB+\angle PBA+\angle APB=180°$ [Angle sum property]

⇒$\angle PAB+\angle PAB=180°–60°=120°$ [Using (1)]

⇒$2\angle PAB=120°$

⇒$\angle PAB=60°$ .............(2)

From (1) and (2)

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

∴ ΔPAB is an equilateral triangle.