Question

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

Answer 1

$PR>PQ$

that means in $△PQR$

$\angle PQR>\angle PRQ$

In $△PSQ$ & $△PSR$

$\angle PRQ+\angle PSR+\angle SPR={180}^o$ (sum of angles of $△={180}^o$)

$\angle PQR+\angle PSQ={180}^o$ (sum of angles of $△={180}^o$)

$\angle SPR=\angle SPQ$ ($PS$ is angle bisector)

then of $\angle PRQ<\angle PQR$

then $\angle PSR>\angle PSQ$