Given: P is a point in the interior of ∠AOB and PL ⊥ OA and PM ⊥ OB such that PL=PM

To prove: ∠POL = ∠POM

Proof:

In ∆OPL and ∆OPM, we have

∠OPM = ∠OPL = 90° …given

OP = OP …common side

PL = PM … given

Thus by Right angle hypotenuse side property of congruence,

∆OPL ≅ ∆OPM

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ ∠POL = ∠POM

Ie. OP is the bisector of ∠AOB