Question

There is a point P outside a circle. From this point, a line segment is drawn joining with centre O of circle. From P, 2 tangents AP and PB are drawn to circle. Prove that ABP is equilateral.

Answer 1

With the information given in the question, it can only be proven that $△ABP$ is isosceles. For it to be equilateral, some more conditions are required. So, let's prove that $△ABP$ is isosceles.

Let us join AO and BO.

$$\begin{gathered} \mathrm{In}△\mathrm{s}\mathrm{AOP}\mathrm{and}\mathrm{BOP}, \\ \angle \mathrm{OAP}=90°=\angle \mathrm{OBP}[\mathrm{since}\mathrm{a}\mathrm{tangent}\mathrm{is}\perp \mathrm{the}\mathrm{line}\mathrm{segment}\mathrm{joing}\mathrm{the}\mathrm{tangent}\mathrm{point}\mathrm{to}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{circle}] \\ \mathrm{OP}=\mathrm{OP}[\mathrm{common}\mathrm{side}\mathrm{in}\mathrm{both}\mathrm{the}△\mathrm{s}] \\ \mathrm{OA}=\mathrm{OB}[\mathrm{both}\mathrm{are}\mathrm{radii}\mathrm{of}\mathrm{the}\mathrm{circle}] \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{by}\mathrm{RHS}\mathrm{congruence},△\mathrm{OAP}\cong △\mathrm{OBP} \\ \Rightarrow \mathrm{AP}=\mathrm{BP}[\mathrm{corresponding}\mathrm{parts}\mathrm{of}\mathrm{congruent}\mathrm{triangles}] \end{gathered}$$

$\Rightarrow △\mathrm{ABP}\mathrm{is}\mathrm{isosceles}.$

Hence Proved.