Question

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

Answer 1

As PR > PQ,

∴ ∠PQR > ∠PRQ (Angle opposite to larger side is larger) ... (1)

PS is 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)

Adding equations (1) and (2), we obtain

∠PQR + ∠QPS > ∠PRQ + ∠RPS

⇒ ∠PSR > ∠PSQ [Using the values of equations (3) and (4)]