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.

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