Question

In fig 6.57 angle q is greater than angle r. pa is the bisector of angle qpr

Answer 1

Answer:

Brainly.in

What is your question?

1

Secondary School Math 5 points

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

Ask for details Follow Report by Sahilazim 24.08.2018

Answers

sarithavalsanpdumbq Ambitious

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)