Question

in question angle Q greater than angle R PA is the bisector of angle QPR and PM perpendicular to QR prove that angle APM =1/2angle Q-angle R

Answer 1

Given: In ΔPQR, ∠ Q > ∠ R. If PA is the bisector of ∠ QPR and PM perp QR.

To Prove: ∠APM =(1/2) ( ∠ Q - ∠ R )

Proof: Since PA is the bisector of ∠P,we have,

∠APQ=(1/2) ∠P....................(i)

In right -angled triangle PMQ,we have,

∠Q+ ∠MPQ=90°

⇒ ∠MPQ= 90°-∠Q...................(ii)

∴∠APM=∠APQ-∠MPQ

1/2 ∠P - (90 - ∠Q) [using (i) and (ii)]

1/2∠P-90+∠Q

1/2∠P - 1/2(∠P + ∠R + ∠Q ) +∠Q [since 90 = 1/2(∠P + ∠R + ∠Q)]

1/2(∠Q -∠R)

Answer 2

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)