Question

In $\Delta PQR,$ If PQ>PR and bisectors of $\angle Q$ and $\angle R$ intersect at S. show that SQ>SR.

Answer 1

In $△PQR,PQ>PR$

$\Rightarrow \angle PRQ>\angle PQR$(Angle opposite to longer side is greater)

$\Rightarrow \frac{1}{2}\angle PRQ>\frac{1}{2}\angle PQR$

$\Rightarrow \angle SRQ>\angle SQR$ since $RS$ and $QS$ are the bisectors of $\angle PRQ$ and $\angle PQR$

In $△SQR,\angle SRQ>\angle SQR$

$\Rightarrow SQ>SR$ since side opposite to greater angle is longer.