Question

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

Answer 1

In $△ADC$,
$\angle ADB=\angle 1+\angle C......\left(i\right)$
In $△ADB$,
$\angle ADC=\angle 2+\angle B.....\left(ii\right)$
But $AC>AB$
$\Rightarrow \angle B>\angle C$
Also given, $\angle 2=\angle 1$ [$AD$ is bisector of $\angle A$]
$\Rightarrow \angle 2+\angle B>\angle 1+\angle C....\left(iii\right)$
From (i), (ii) and (iii)
$\Rightarrow \angle ADC>\angle ADB$