It is given that PR > PQ and PS bisects QPR

Now we will have to prove that angle PSR is smaller than PSQ i.e. PSR > PSQ

Proof:

QPS = RPS — (ii) (As PS bisects ∠QPR)

PQR > PRQ — (i) (Since PR > PQ as angle opposite to the larger side is always larger)

PSR = PQR + QPS — (iii) (Since the exterior angle of a triangle equals to the sum of opposite interior angles)

PSQ = PRQ + RPS — (iv) (As the exterior angle of a triangle equals to the sum of opposite interior angles)

By adding (i) and (ii)

PQR + QPS > PRQ + RPS

Thus, from (i), (ii), (iii) and (iv), we get

PSR > PSQ