Ex.7.4 Q5 Triangles Solution - NCERT Maths Class 9

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Question

In the given figure, \(PR \gt PQ\) and \(PS\)

bisects \(\angle QPR\). Prove that

\(\angle PSR \gt \angle PSQ\).

 Video Solution
Triangles
Ex 7.4 | Question 5

Text Solution

What is Known?

\(\text{PR  PQ}\) and PS bisects \(\angle \text{QPR}\)

To prove:

\[\angle \text{PSR }>\angle \text{PSQ}\text{.}\]

Reasoning:

We can use exterior angle sum property to find the required inequality.

Steps:

As \(PR \gt PQ\),

\[\begin{align} & \angle PQR \gt \angle PRQ\\&\left(\begin{array}{} \text {Angle opposite to larger}\\\text{ side is larger}) \ldots (1)\end{array}\right) \\\\ & PS \text { is the bisector of } \angle QPR \\ & \angle QPS = \angle RPS \ldots (2) \\\\ & \angle PSR \text{ is the exterior angle of } PQS \\ & \angle PSR = \angle PQR + \angle QPS \ldots (3)\\\\ & \angle PSQ \text{ is the exterior angle of } PRS \\ & \angle PSQ = \angle PRQ + \angle RPS \ldots (4)\\ \end{align}\]

Adding Equations (\(1\)) and (\(2\)), we obtain

\[\begin{align} &\angle PQR + \angle QPS \gt \angle PRQ + \angle RPS \\ & \angle PSR \gt \angle PSQ \\& \begin{bmatrix}\text{Using the values of}\\ \text{Equations (2),(3) and (4)}\end{bmatrix}\end{align}\]

 Video Solution
Triangles
Ex 7.4 | Question 5
  
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