# In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

**Solution:**

Given: PR > PQ and PS bisects ∠QPR

To prove: ∠PSR > ∠PSQ

As PR > PQ, ∠PQS > ∠PRS (Angle opposite to larger side is larger ) ...(1)

PS is the angle bisector of ∠QPR.

∠QPS = ∠RPS ...(2) [Since,PS bisects ∠QPR]

∠PSR is the exterior angle of ΔPQS.

∠PSR = ∠PQS + ∠QPS ...(3) [Using exterior angle sum property]

∠PSQ is the exterior angle of ΔPRS.

∠PSQ = ∠PRS + ∠RPS ...(4) [Using exterior angle sum property]

From Equations (1) and (2), we obtain

∠PQS + ∠QPS > ∠PRS + ∠RPS

∠PSR > ∠PSQ [From Equations (3) and (4)]

**Video Solution:**

## In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

### NCERT Maths Solutions Class 9 - Chapter 7 Exercise 7.4 Question 5:

**Summary:**

If in the given figure, PR > PQ and PS bisects ∠QPR, we have proved that ∠PSR > ∠PSQ.