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 in the opposite of the longer side is greater) .............................(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))

$\therefore 180-(\angle PSQ+\angle QPS)>180-(\angle PSR+\angle RPS)$ ($\angle QPS=\angle PRQ)$

$\Rightarrow \angle PSR>\angle PSQ$ $\left[henceproved\right]$