Question

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

Answer 1

In $\Delta ABC$
$AD$ is the bisector of $\angle BAC$
$\angle 1=\angle 2$ …(i)
Also in $\Delta ADC$
$\angle 3=\angle 2+\angle C$ (ext. angle is equal to sum of opposite interior angles)
$\angle 3>\angle 2$ (ext. angle is greater than one of the interior angles)
But $\angle 1=\angle 2$
$\angle 3>\angle 1$
$AB>BD$
Hence proved.