Question

In ΔPQR, angle P=90° and M is a point on QR such that PM⊥QR. The internal bisector of ∠MQP meets MP at A and the internal bisector of ∠MPR meets MR at B. Prove that AB is parallel to PR.

Answer 1

Answer:

∵∠QPM=∠RPM

In ΔPNQ

∠QPN=90

∘

−∠Q

In ΔPNR

∠RPN=90

∘

−∠R

∠QPM=∠RPM

⇒∠QPN+∠MPN=∠RPN−∠MPN

⇒2∠MPN=∠RPN−∠QPN

⇒2∠MPN=(90

∘

−∠R)−(90

∘

−∠)

⇒∠MPN=

2

1

(∠Q−∠R) (proved)

Answer 2

Answer:

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