In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.
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)]
Concept: Inequalities in a Triangle
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