Question

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

Answer 1

Consider an angle ABC and BP be one of the arm within the angle.

Draw perpendiculars PN and PM on the arms BC and BA. 

In Δ BPM and Δ BPN, 

∠ BMP = ∠ BNP = 90° [given] 

BP = BP [Common side] 

MP = NP [given] 

By RHS congruence criterion: ΔBPM≅ΔBPN 

So, ∠ MBP = ∠ NBP [ By CPCT] 

BP is the angular bisector of ∠ABC. 

Hence proved