Question

From a point $P$, two tangents $PA$ and $PB$ are drawn to a circle $C(O,r)$. If $OP=2r$, show that $△APB$ is equilateral.

Answer 1

In$\Delta OAP:$

$OA=r$

$OP=2r$

$\Rightarrow P{A}^2=O{P}^2-O{A}^2=4r^2-r^2=3r^2$

$\Rightarrow PA=\sqrt{3}r\therefore PA=PB=\sqrt{3}r$

$(\because$Tangents from an external point are equal in length $)$

Now, $\tan \left(\angle APO\right)=\frac{OA}{AP}=\frac{r}{r\sqrt{3}}=\frac{1}{\sqrt{3}}\angle APO=30°$

Similarly

$\angle BPO=30°\therefore \angle APB=60°$

In $\Delta ABP,$

Using cosine rule $:$

$\cos \left(\angle APB\right)=\frac{A{P}^2+B{P}^2-A{B}^2}{2AP.BP}\Rightarrow \cos \left(60°\right)=\frac{{\left(\sqrt{3}r\right)}^2+{\left(\sqrt{3}r\right)}^2-A{B}^2}{2{\left(\sqrt{3}r\right)}^2}\Rightarrow \frac{1}{2}=\frac{6r^2-A{B}^2}{2\left(3r^2\right)}\Rightarrow 3r^2=6r^2-A{B}^2\Rightarrow AB=\sqrt{3}r\therefore AB=AP=BP=\sqrt{3}r$

Hence, $ABP$ is an equilateral triangle