Question

In$∆$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
Now, the angle opposite to the greater side is greater than angle opposite to the smaller side.
∴ ∠R > ∠Q
Dividing both sides by 2, we get
$\frac{1}{2}\angle \mathrm{R}>\frac{{\displaystyle 1}}{{\displaystyle 2}}\angle \mathrm{Q}$
⇒ ∠SRQ > ∠SQR
⇒ SQ > SR (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence proved.