S is a point on the side QR of triangle PQR such that PS is the bisector of angle QPR. then: A. QS = SR B. QP > QS C. QS > QP D. SR > RP

  B. QP > QS

According to the given condition, ∠QPS=∠RPS

By external angle theorem, 

∠PSQ=∠RPS+∠PRS

⇒∠PSQ=∠QPS+∠PRS

⇒∠PSQ>∠QPS

In a triangle, the side opposite to the greater angle is greater. Therefore,

⇒QP>QS