In given fig, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

In ΔPQR, PR > PQ (Given) ⇒ ∠PQR > ∠PRQ.. (1) (Angle opposite to side of greater length is greater)
PS is the bisector of ∠P, so ∠x = ∠y
Adding ∠x in (1) ⇒ ∠PQR + ∠x > ∠PRQ + ∠x
⇒ ∠PQR + ∠x > ∠PRQ + ∠y (2)
In ΔPQS, ∠PQS + ∠x + ∠PSQ = 180° (Angle sum property of triangle)
∴ ∠PQS + ∠x = 180° - ∠PSQ. (3)
In ΔPSR, ∠PRS + ∠y + ∠PSR = 180° (Angle sum property of triangle)
∠PRS + ∠y = 180° - ∠PSR
Using equation (1), (2), (3) we get
180° - ∠PSQ > 180° - ∠PSR.. (4)
⇒ -∠PSQ > -∠PSR
⇒ ∠PSQ < ∠PSR
So, ∠PSR > ∠PSQ (hence proved)