In the figure,$AD$ is the bisector of $\angle BAC$. Prove that $AB > BD$.

(N/A) Let $\angle BAD = \angle 1$ and $\angle CAD = \angle 2$. Since $AD$ is the bisector of $\angle BAC$,we have $\angle 1 = \angle 2$.
In $\triangle ACD$,$\angle ADC$ is an exterior angle at vertex $D$.
We know that an exterior angle of a triangle is equal to the sum of the two interior opposite angles.
Thus,$\angle ADC = \angle CAD + \angle ACD = \angle 2 + \angle ACD$.
Since $\angle ACD > 0$,it follows that $\angle ADC > \angle 2$.
Substituting $\angle 2 = \angle 1$,we get $\angle ADC > \angle 1$.
Now,consider $\triangle ABD$. In this triangle,the angle opposite to side $AB$ is $\angle ADB$ (which is $\angle ADC$) and the angle opposite to side $BD$ is $\angle BAD$ (which is $\angle 1$).
Since $\angle ADC > \angle 1$,the side opposite to the greater angle is longer.
Therefore,$AB > BD$.