In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.


As PR $>P Q$

$\therefore \angle \mathrm{PQR}>\angle \mathrm{PRQ}$ (Angle opposite to larger side is larger) ... (1)

$P S$ is the bisector of $\angle Q P R$.

$\therefore \angle Q P S=\angle R P S \ldots(2)$

$\angle P S R$ is the exterior angle of $\triangle P Q S$.

$\therefore \angle \mathrm{PSR}=\angle \mathrm{PQR}+\angle \mathrm{QPS} \ldots(3)$

$\angle P S Q$ is the exterior angle of $\triangle P R S$.

$\therefore \angle \mathrm{PSQ}=\angle \mathrm{PRQ}+\angle \mathrm{RPS} \ldots(4)$

Adding equations $(1)$ and $(2)$, we obtain

$\angle \mathrm{PQR}+\angle \mathrm{QPS}>\angle \mathrm{PRQ}+\angle \mathrm{RPS}$

$\Rightarrow \angle \mathrm{PSR}>\angle \mathrm{PSQ}[U$ sing the values of equations $(3)$ and $(4)]$

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