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 the figure.

The angle opposite to a larger side is larger than the angle opposite to a smaller side.

It is given

$AB<AC$

$BP$ is the bisector of $\angle B$

So,

$\angle 1=\angle 2=\frac{1}{2}\angle B$

$CP$ is bisector of $\angle C$

$\angle 3=\angle 4=\frac{1}{2}\angle C$

Angle opposite to $AB$ is $\angle C$ and angle opposite to $AC$ is $\angle B.$

Now, from the property

As $AB<AC$

So,

$\angle C<\angle B$

$\Rightarrow \frac{1}{2}\angle C<\frac{1}{2}\angle B$

$\Rightarrow \angle 4<\angle 2$

Side opposite to $\angle 2$ is $PC$ and side opposite to $\angle 4$ is $PB.$

Hence,

$\angle 4<\angle 2$

$\Rightarrow PB<PC$

Hence, Proved.