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# Q is a point on the side SR of a ∆ PSR such that PQ = PR. Prove that PS > PQ.

**Solution:**

Given, PSR is a triangle.

Q is a point on the side SR of the triangle

PQ = PR

We have to show that PS > PQ

We know that the angles opposite to the equal sides are equal.

∠PRQ = ∠PQR --------------- (1)

We know that the exterior angle of a triangle is greater than each of the opposite interior angles.

∠PQR > ∠S ------------------- (2)

From (1) and (2),

∠PRQ > ∠S

We know that in a triangle side opposite to greater angle is longer.

So, PS > PR

Given, PQ = PR

Therefore, PS > PQ

**✦ Try This:** In the given figure, if ∠ADE = ∠B, show that ΔADE~ΔABC. If AD=3.8cm, AE=3.6cm ,BE=2.1cm and BC=4.2cm, find DE.

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 7

**NCERT Exemplar Class 9 Maths Exercise 7.3 Problem 5**

## Q is a point on the side SR of a ∆ PSR such that PQ = PR. Prove that PS > PQ

**Summary:**

Q is a point on the side SR of a ∆ PSR such that PQ = PR. It is proven that PS > PQ

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