Question
In a $\triangle\text{ ABC},\text{ AD}$ bisects $\angle\text{A}$ and $\angle\text{C} > \angle\text{B}.$. Prove that $\angle\text{ADB} > \angle\text{ADC}.$
✓
Answer
$\therefore\angle\text{C}>\angle\text{B}$ [Given]$\Rightarrow\angle\text{C}+\text{x} > \angle\text{B}+\text{x}$ [Adding x on both sides]
$\Rightarrow180^\circ-\angle\text{ADC}>180^\circ-\angle\text{ADB}$
$\Rightarrow-\angle\text{ADC}>-\angle\text{ADB}$
$\Rightarrow\angle\text{ADB}>\angle\text{ADC}$
hence proved.
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