In Figure given below, ∠ Q > ∠ R, PA is the bisector of ∠ QPR and PM
Your question and diagram seems to be still wrong. So, if your question is :
In ΔPQR, ∠ Q > ∠ R. If PA is the bisector of ∠ QPR and PM QR.
Prove that ∠APM =(1/2) ( ∠ Q - ∠ R )
Then the Solution is as follows:
Given: In ΔPQR, ∠ Q > ∠ R. If PA is the bisector of ∠ QPR and PM 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
Hence, the result.