Question -
Answer -
AsPR > PQ,
∴ ∠PQR > ∠PRQ(Angle opposite to larger side is larger) … (1)
PSis the bisector of ∠QPR.
∴∠QPS = ∠RPS … (2)
∠PSR is the exterior angle of ΔPQS.
∴ ∠PSR = ∠PQR+ ∠QPS … (3)
∠PSQ is the exterior angle of ΔPRS.
∴ ∠PSQ = ∠PRQ+ ∠RPS … (4)
Addingequations (1) and (2), we obtain
∠PQR + ∠QPS > ∠PRQ+ ∠RPS
⇒ ∠PSR > ∠PSQ[Using the values of equations (3) and (4)]