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

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,

$\angle PSR > \angle PSQ$.

Tutorialspoint

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Updated on: 10-Oct-2022

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 $','sharer','toolbar=0,status=0,width=580,height=325');" href="javascript: void(0)" title="Share on Facebook"> PQ$ 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 $','sharer','toolbar=0,status=0,width=580,height=325');" href="javascript: void(0)" title="Share on LinkedIn">

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