Given,

In ΔPQR,

PS bisects ∠QPR and QS = SR

To prove:

PQ = PR

In ΔPQS and ΔPRS

QS = SR [Given]

∠QPS = ∠RPS [Given]

PS = PS [Common]

ΔPQS is congruent to ΔPRS [S.A.S]

⸫ PQ = PR [C.P.C.T.C]

Hence Proved.