IN triangle.................

construction: draw a line thru' R parallel to PS meeting QP in T(say)

now, consider triangle QTR,

 we have

 PS II TR

so by BPT,

QP/PT=QS/SR..(i)

angle QPS= angle RPS.. ..(ii) since PS is the bisector.

angle QPS= angle PTR...(iii) corr angles with SPII TR

angle RPS= angle PRT... ..(iv) alt interior angles

so by (ii) we get,

angel PRT= angle PTR

So,

PT=PR... sides opp equal angles are equal.

hence from (i0 we now get,

QP/PR=QS/SR