Question

In $△PQR$, if $PQ>PR$ and bisectors of $\angle Q$ and $\angle R$ intersect at $S$. Show that $SQ>SR$.

Answer 1

Given: In $APQR,PQ>PR$ and bisectors of $\angle Q$ and $\angle R$ intersect at $S$.
To prove: $SQ>SR$
Solution:
Proof:
$\angle SQR=\frac{1}{2}\angle PQR$....(i) [Ray QS bisects $\angle PQR]$
$\angle SRQ=\frac{1}{2}\angle PRQ$....(ii) [Ray RS bisects $\angle PRQ$ ]
In $△PQR$,
$PQ>PR$ [Given]
$\therefore \angle R>\angle Q$ [Angle opposite to greater side is greater.]
$\therefore \frac{1}{2}\left(\angle R\right)>\frac{1}{2}\left(\angle Q\right)$ [Multiplying both sides by $\frac{1}{2}$ ]
$\therefore \angle SRQ>\angle SQR$....(iii) [From (i) and (ii)]
In $△SQR$,
$\angle SRQ>\angle SQR$ [From (iii)]
$\therefore SQ>SR$ [Side opposite to greater angle is greater]