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

Proof:

Given that perpendiculars from any point within an angle on its arms are congruent.

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

Now, let us consider an ∠ABC and let BP be one of the arms within the angle.

Draw perpendicular PN and PM on the arms BC and BA

Such that,

They meet BC and BA in N and M respectively.

Now, in ∆BPM and ∆BPN

We have,

∠BMP = ∠BNP = 90^o (Given)

BP = BP (Common)

MP = NP (Given)

So, by RHS congruence rule, we have

∆BPM ≅BPN

∠MBP = ∠NBP (By c.p.c.t)

BP is the angular bisector of ∠ABC

Hence, proved