Question

In the adjoining figure, P is a point in the interior of ∠AOB. If PL ⊥ OA and PM ⊥ OB such that PL = PM, show that OP is the bisector of ∠AOB.

Answer 1

$$\begin{gathered} \mathrm{In}△\mathrm{OMP}\mathrm{and}△\mathrm{OLP},\mathrm{we}\mathrm{have}: \\ \mathrm{OP}=\mathrm{OP}\left(\mathrm{Common}\right) \\ \mathrm{PL}=\mathrm{PM}\left(\mathrm{Given}\right) \\ \angle \mathrm{BMP}=\angle \mathrm{PLA}=90° \\ \Rightarrow △\mathrm{OMP}\cong ∆\mathrm{OPL}(\mathrm{RHS}\mathrm{criterion}) \end{gathered}$$

i.e., $\angle POM=\angle POL$

Thus, OP is the bisector of $\angle AOB$.