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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