‘D’ is a point on the side BC of a triangle ABC such that AD bisects $\angle BAC$ is

Given in $\Delta ABC$ , AD bisects $\angle BAC$

Let $\angle BAD=x$

$\Rightarrow \angle CAD=x$

$\angle ABC=y$

$\angle ACB=z$

We know that exterior angle is equal to sum of its opposite interior angles

$\angle ADB=x+z$

$\angle ADC=x+y$

$\Delta ABD,\angle BDA=x+z$

$\angle BAD=x$

$\therefore x+z>x$

$\Rightarrow x+z>x$

$\Rightarrow AB>BD$