Question

In adjacent figure, $PR>PQ$ and $PS$ bisects $\angle QPR$. Prove that $\angle PSR>\angle PSQ$.

Answer 1

In the given triangle $PQR,PR>PQ$ and $PS$ bisects $\angle QPR$
In $\Delta PQR$
$PR>PQ$ ....(Given )
$\therefore \angle PQR>\angle PRQ$ (Angle opposite to longer side is larger) .... (1)
In $\Delta PQS$ and $\Delta RPS$
$\angle PQR=\angle PQS=180°-(\angle PSQ+\angle QPS)$
$\angle PRQ=\angle PRS=180°-(\angle PSR+\angle RPS)$
Given $PS$ is bisector of $\angle QPR$
Then $\angle QPS=\angle RPS$ .... (2)
$\angle PQR>\angle PRQ$ (As per (1))
$\therefore 180°-(\angle PSQ+\angle QPS)>180°-(\angle PSR+\angle RPS)$
$\Rightarrow \angle PSR>\angle PSQ$ $(\because \angle QPS=\angle RPS)$