Given that, if perpendicular from any point within, an angle on its arms is congruent, prove that it lies on the bisector of that angle
Let us consider an angle 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 ΔBPM and ΔBPN
∠BMP =∠BNP = 90° [given]
BP = BP [Common side]
And MP = NP [given]
So, by RHS congruence criterion, we have
ΔBPM ≅ ΔBPN
∠MBP =∠NBP [ ∵ Corresponding parts of congruent triangles are equal]
⇒ BP is the angular bisector of ∠ABC.
∴ Hence proved