In a $\triangle ABC, AD$ bisects $\angle A$ and $\angle C > \angle B$. Prove that $\angle ADB > \angle ADC$.
Given:
In a $\triangle ABC, AD$ bisects $\angle A$ and $\angle C > \angle B$.
To do:
We have to prove that $\angle ADB > \angle ADC$.
Solution:
In $\triangle ABC, AD$ is the bisector of $\angle A$.
This implies,
$\angle 1 = \angle 2$
In $\triangle ADC$,
$\angle ADB = \angle l+ \angle C$
$\angle C = \angle ADB - \angle 1$.....…(i)
Similarly,
In $\triangle ABD$,
$\angle ADC = \angle 2 + \angle B$
$\angle B = \angle ADC - \angle 2$......…(ii)
$\angle C > \angle B$
From (i) and (ii), we get,
$(\angle ADB - \angle 1) > (\angle ADC - \angle 2)$
$\angle 1 = \angle 2$
Therefore,
$\angle ADB > \angle ADC$.
Hence proved.
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