Question

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

Answer 1

Given that-

$\Delta PQR,PQ>PR$

$\angle Q\&\angle R$ have bisector intersected at $S$

To find-

To prove $SQ>SR$

$PQ>PR$

$\therefore \angle PQR>\angle PRQ$

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

$\therefore \angle SRQ>\angle SQR$ [$\because QS$ and $RS$ bisectors of $\angle Q$ and $LR$ respectively]

$\Rightarrow SQ>SR$ (Since, $\angle SRQ$ is larger and is formed by)

Hence, proved $SQ$