# In Fig. 6.56, PS is the bisector of ∠QPR of ΔPQR. Prove that QS/SR = PQ/PR

**Solution:**

As we know, if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. (Basic Proportionality Theorem)

Draw a line parallel to PS, through R, which intersect QP produced at T.

PS || RT

In ΔQPR

∠QPS = ∠SPR (Since PS is the bisector of ∠QPR) (i)

But ∠PRT = ∠SPR (alternate interior angles) (ii)

∠QPS = ∠PTR (Corresponding angles) (iii)

From (i), (ii), and (iii)

∠PTR = ∠PRT

PR = PT................ (iv)

(Since in a triangle sides opposite to equal angles are equal)

In ΔQRT, PS || RT

QS/SR = QP/PT [Basic Proportionality Theorem]

QS/SR = QP/PR [from (iv)]

**Video Solution:**

## In Fig. 6.56, PS is the bisector of ∠QPR of ΔPQR. Prove that QS/SR = PQ/PR

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.6 Question 1:

In Fig. 6.56, PS is the bisector of ∠QPR of ΔPQR. Prove that QS/SR = PQ/PR

In the given figure, PS is the bisector of ∠QPR of ΔPQR. Hence prove that QS/SR = PQ/PR