"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

= ∠Q/2-∠R/2

=1/2(∠Q-∠R)

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