Question

angle q > angle r and m is a point on qr such that pm is the bisetor of angle qpr . if perpendicular from pq on qr meets qp at n .prove that angle npm = 1/2 (angle q - angle r )

Answer 1

Step-by-step explanation:

Given: angle Q is greater than angle R and PA is the bisector of angle QPR and PM perpendicular QR

To Prove: angle APM = 1/2 (Q - R)

Proof: In triangle PQM

=> PMQ + MPQ + Q = 180 [Angle sum property of a triangle[

=> 90 + MPQ + Q = 180 (PMQ = 90)

=> Q = 90 - MPQ

=> 90 - MPQ = Q _____(1)

In triangle PMR

=> PMR + PRM + R = 180

=> 90 + MPR + R = 180

=> R = 90 - MPR

=> 90 - MPR = R _____(2)

Subtracting (1) and (2), we get

=> (90 - MPQ) - (90 - MPR) = Q - R

=> MPR - MPQ = Q - R

=> (RPA +APM) - (QPA - APM) = Q - R

=> QPA + APM -QPA + APM = Q - R [ As PA is bisector of QPR so, RPA=QPA]

=> 2 APM = Q - R

=> APM = 1/2 (Q - R)