Question

In $\Delta$ABC, AB$<$AC, PB and PC are the bisectors of $\angle$B and $\angle$C. Prove that PB$<$ PC.

Answer 1

In $△ABC,$

$AB>AC$

$\Rightarrow \angle ABC<\angle ACB$

$\therefore {180}^{\circ }-\angle ABC>{180}^{\circ }-\angle ACB$

$\Rightarrow \frac{{180}^{\circ }-\angle ABC}{2}>\frac{{180}^{\circ }-\angle ACB}{2}$

$\Rightarrow {90}^{\circ }-\frac{1}{2}\angle ABC>{90}^{\circ }-\frac{1}{2}\angle ACB$

$\Rightarrow \angle CBP>\angle BCP$ since $BP$ is a bisector of $\angle CBD$ and $CP$ is bisector of $\angle BCE$

$\Rightarrow PC>PB$ since side opposite to greater angle is greater.