Question

In triangle $ABC$, side $AC$ is greater then side $AB$. If the internal bisector of angle $A$ meets the opposite side at point $D$, prove that
$\angle ADC$ is greater then $\angle ADB$.

Answer 1

Since $AC>AB$

So, $\angle B>\angle C$

As angle opposite to the greater side is greater.

Now,

In $△ADC$ and $△ADB$

$\angle ADC+\angle C+\angle DAC={180}^{\circ }=\angle ADB+\angle B+\angle DAB$

Since $\angle DAC=\angle ADB$

$\angle ADC+\angle C=\angle B+\angle ADB$

But, $\angle B>\angle C$

So, to hold equality,

$\angle ADC>\angle ADB$