Question

In fig., OP is equal to diameter of the circle. Prove that ΔABP is an equilateral triangle.

Answer 1

Join AB

Sin $\angle$APO = $\frac{OA}{OP}$ = $\frac{r}{2r}$

Sin $\angle$APO = $\frac{1}{2}$

$\angle$APO = ${30}^{\circ }$

Similarly

$\angle$BPO = ${30}^{\circ }$

$\therefore$ $\angle$APB = $\angle$APO + $\angle$BPO = ${30}^{\circ }$ + ${30}^{\circ }$ = ${60}^{\circ }$

As the lengths of tangents drawn from an external point to a circle are equal,

PA = PB

$\angle$PAB = $\angle$PBA

In $\Delta$PAB, $\angle$ABP + $\angle$BAP + $\angle$APB = ${180}^{\circ }$

2$\angle$ABP + ${60}^{\circ }$ = ${180}^{\circ }$

$\angle$ABP =${60}^{\circ }$

Therefore $\angle$BAP = ${60}^{\circ }$

$\Delta$PAB is an equilateral triangle.