Question
In $\triangle ABC, \text{AB} > \text{AC}$ and $D$ is a point inside $\text{BC}$.Show that$: \text{AB} > \text{AD}.$
✓
Answer
Given that, $\text{AB}>\text{AC}$
$\Rightarrow \angle C>\angle B ...(i$)
Also in $\triangle ADC$
$ \angle ADB =\angle DAC +\angle C \ldots[$ Exterior angle$]$
$ \Rightarrow \angle ADB >\angle C$
$ \Rightarrow \angle ADB >\angle C >\angle B \ldots[$From $(i) ]$
$ \Rightarrow \angle ADB >\angle B$
$ \Rightarrow \text{AB} >\text{AD} .$
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