Question

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

Answer 1

In $△AOP$

$\cos \theta (\theta =\angle AOP)=\frac{AO}{OP}=\frac{r}{2r}=\frac{1}{2}$

$\Rightarrow$ $\theta ={60}^o$

$\Rightarrow$ $\angle AOP=180-90-60={30}^o$

Similarly $\angle BPO={30}^o$ $

$\Rightarrow$ $\angle APB={60}^o$

Now, $PA=PB$ (tangents from same point are equal)

$\Rightarrow$ $\angle PAB=\angle PBA$

But $\angle PAB+\angle PBA+{60}^o={180}^o$

$\Rightarrow$ $\angle PAB=\angle PBA={60}^o$

$\Rightarrow$ $△ABP$ is equilateral triagnle