In ΔPQR, the bisector of ∠QPR intersects QR at S and the circumcircle of ΔPQR at T. If PQ : PS = 3 : 5, then PT : PR is ?

3 : 5 5 : 3 1 : 1 2 : 5

3 : 5

Here, ∠PTQ = PRQ,   (angle made by same chord PQ) ⇒ ΔPSR ∼ ΔPQT ⇒\(\frac{PS}{PR}\) = \(\frac{PQ}{PT}\) ⇒ \(\frac{5}{PR}\) = \(\frac{3}{PT}\) = \(\frac{PT}{PR}\) = \(\frac{3}{5}\)