# 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

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

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