In Fig 7.51, PR $ > $ PQ and $ \mathrm{PS} $ bisects $ \angle \mathrm{QPR} $. Prove that $ \angle \mathrm{PSR}>\angle \mathrm{PSQ} $.
"

AcademicMathematicsNCERTClass 9

Given:

$PR > PQ$ and $PS$ bisects $\angle QPR$.

To do:

We have to prove that $\angle PSR > \angle PSQ$.

Solution:

Let us consider $\triangle PQR$

We have,

$PR > PQ$

We know that,

The angle opposite the longer side will always be larger.

This implies,

$\angle PQR > \angle PRQ$...(i)

Since we have, $PS$ bisects $\angle QPR$

We get,

$\angle QPS=\angle RPS$...(ii)

We also that,

The sum of opposite interior angles of a triangle is equal to the exterior angle,

This implies,

In $\triangle PSR$,

$\angle PSR=\angle PQR+\angle QPS$...(iii)

In $\triangle PSQ$,

$\angle PSQ=\angle PRQ+\angle RPS$...(iv)

By adding (i) and (ii)

We get,

$\angle PQR+\angle QPS > \angle PRQ+\angle RPS$

Therefore,

From (i), (ii), (iii), (iv)

We get,

$\angle PSR > \angle PSQ$.

raja
Updated on 10-Oct-2022 13:41:24

Advertisements