If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.

Given:

The perpendiculars from any point within an angle on its arms are congruent

To do:

We have to prove that it lies on the bisector of that angle.

Solution:

Let a point $P$ lies in $\angle ABC$, $PL \perp BA$, $PM \perp BC$ and $PL = PM$.

$PB$ is joined.

In $\triangle PLB$ and $\triangle PMB$

$PL = PM$

$PB = PB$             (Common)

Therefore, by RHS axiom,

$\triangle PLB \cong \triangle PMB$

This implies,

$\angle PBL = \angle PBM$             (CPCT)

Therefore,

$PB$ is the bisector of $\angle ABC$.

Hence proved.

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Updated on: 10-Oct-2022

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