Question

In figure,angle Q biger with angle R , PA IS THE BISECTOR OF ANGLE QPR AND PM PERPENDICULAR IN QR,PROVE THAT ANGLE APM =1/2(ANGLE Q - ANGLE R)

Answer 1

Step-by-step explanation:

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)