In Δ PQR, If PQ > PR and bisectors of ∠ Q and ∠ R intersect at S. Show that SQ > SR.
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
`1/2 angle "R" > 1/2 angle "Q"`
⇒ ∠SRQ > ∠SQR
⇒ SQ > SR (The sides opposite to the greater angle is greater than side opposite to the smaller angle)