Question

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

Answer 1

Since exterior angle of a triangle is greater than either of the interior opposite angles, therefore, in $△ACD$,
Ext $\angle ADB>\angle DAC⟹ADB>\angle BAD$
[$\therefore AD$ is the bisector of $△BAC$, so $\angle DAC=\angle BAD$]
Now, in $△ABD$, we have
$\angle ADB>\angle BAD$
Hence, $AB>BD$. [$\therefore$ In a triangle, side opposite to greater angle is longer]