Question

In the following figure, ∠Q > ∠R, PA is the bisector of ∠QPR and PM ⊥ QR. Prove that ∠APM = `1/2` (∠Q – ∠R).

Answer 1

Given in triangle PQR, ∠Q > ∠R, PA is the bisector of ∠QPR and PM ⊥ QR.

To prove that ∠APM = `1/2`(∠Q – ∠R)

Proof: PA is the bisector of ∠QPR.

So, ∠QPA = ∠APR

In angle PQM, ∠Q + ∠PMQ + ∠QPM = 180°  ...(I) [Angle sum property of a triangle]

∠Q + 90° + ∠QPM = 180°   ...[∠PMR = 90°]

∠Q = 90° – ∠QPM  ...(II)

In triangle PMR, ∠PMR + ∠R + ∠RPM = 180°   ...[Angle sum property of a triangle]

90° + ∠R + ∠RPM = 180°   ...[∠PMR = 90°]

∠R = 180° – 90° – ∠RPM

∠R = 90° – ∠RPM  ...(III)

Subtracting equation (III) from equation (II), we get

∠Q – ∠R = (90° – ∠APM) – (90° – ∠RPM) 

∠Q – ∠R = ∠RPM – ∠QPM

∠Q – ∠R = (∠RPA + ∠APM) – (∠QPA – ∠APM)  ...(IV)

∠Q – ∠R = ∠QPA + ∠APM – ∠QPA + ∠APM  ...[As, ∠RPA = ∠QPA]

∠Q – ∠R = 2∠APM

∠APM = `1/2`(∠Q – ∠R)

Hence proved.