Question

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

Answer 1

In $\Delta PQR$,

$PR>PQ$ (Given)

$\Rightarrow \angle PQR>\angle PRQ$ ...(1)

(Angle opposite to side of greater length is greater

$PS$ is the bisector of $\angle P,$ so$\angle x=\angle y$

Adding $\angle x$ in (1)

$\Rightarrow \angle PQR+\angle x>\angle PRQ+\angle x$

$\Rightarrow \angle PQR+\angle x>\angle PRQ+\angle y$ .... (2)

In $\Delta PQS$,

$\angle PQS+\angle x+\angle PSQ={180}^{\circ }$

(Angle sum property of triangle)

$\therefore \angle PQS+\angle x={180}^{\circ }-\angle PSQ$ ..... (3)

In $\Delta PSR$ ,

$\angle PRS+\angle y+\angle PSR={180}^{\circ }$

(Angle sum property of triangle)

$\angle PRS+\angle y={180}^{\circ }-\angle PSR$

Using equation (1), (2), (3) we get

${180}^{\circ }–\angle PSQ>{180}^{\circ }-\angle PSR$ ...(4)

$\Rightarrow -\angle PSQ>-\angle PSR$

$\Rightarrow \angle PSQ<\angle PSR$

So,

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