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

**Solution:**

According to Basic Proportionality Theorem, 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.

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

From the figure, 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 equations (i), (ii), and (iii)

∠PTR = ∠PRT

Thus, PR = PT (Since in a triangle sides opposite to equal angles are equal) ....... (iv)

In ΔQRT, PS || RT

QS/SR = QP/PT (Basic Proportionality Theorem)

Thus, QS/SR = QP/PR [From (iv)]

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 6

**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

**Summary:**

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

**☛ Related Questions:**

- In Fig. 6.57, D is a point on hypotenuse AC of ΔABC , such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that: (i) DM² = DN.MC (ii) DN² = DM.AN.
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