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 arm 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

We have,

∠BMP = ∠BNP = 90^o (Given)

BP = BP (Common)

MP = NP (Given)

So, by RHS congruence rule, we have

∠MBP = ∠NBP (By c.p.c.t)

BP is the angular bisector of ∠ABC

Hence, proved