In fig., PS is the bisector of ∠QPR of ΔPQR. Prove that QS/SR = PQ/PR.

Given: ∠QPS = ∠RPS
To Prove: QS/SR = PQ/PR
Construction: Extend RP to T and Join QT such that TQ || PS
Proof: Since, QT || PS
∴ ∠TQP = ∠QPS (Alternate Angles)
Also, ∠QTP = ∠RPS = ∠QPS (Corresponding Angles and PS is the bisector of ∠QPR of ΔPQR)
∴ ∠TQP = ∠QTP
∴ TP = QP.. (1)
Since, QT || PS, by basic proportionality theorem,
∴ QS/SR = TP/PR
∴ QS/SR = PQ/PR (From 1)
Hence proved