Question

Question 5
In the given figure, PR > PQ and PS bisects $\angle$QPR. Prove that $\angle$PSR > $\angle$PSQ
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Answer 1

PR>PQ (Given)
$\angle$PQR > $\angle$PRQ.........(1) (In any triangle, the angle opposite to the longer side is larger.)
We also have $\angle PQR+\angle QPS+\angle PSQ={180}^{\circ }$ (Angle sum property of triangle)
$\therefore \angle PQR={180}^{\circ }-\angle QPS-\angle PSQ$.....(2)
And, $\angle PRQ+\angle RPS+\angle PSR={180}^{\circ }$ (Angle sum property of triangle)
$\therefore \angle PRQ={180}^{\circ }-\angle PSR-\angle RPS$ .........(3)
Substituting (2) and (3) in (1), we get
${180}^{\circ }-\angle QPS-\angle PSQ$ > ${180}^{\circ }-\angle PSR-\angle RPS$
We also have $\angle QPS=\angle RPS$, using this in the above equation, we get
$-\angle PSQ$ > $-\angle PSR$
$\Rightarrow \angle PSQ$ < $\angle PSR$