"Angle Q is greater than angle R and M is a point on QR such that PM is the bisector of angle QPR If the perpendicular from P on QR meets at N, prove that angle MPN =1/2(angle Q-angle R)
∠MPN = ∠P-∠QPN-∠MPR
=∠P-∠QPN-1/2 ∠P(Because MP is the angular bisector of ∠P).
=1/2 ∠P -∠QPN
=1/2 ∠P -(90 - ∠Q) (because triangle PQN is right angled).
= 1/2 (180-∠Q-∠R) - (90-∠Q) (because the sum of angles in triangle PQR is 180)
=90 - ∠Q/2 - ∠R/2 -90 +∠Q