Question
Prove that a $\triangle ABC$ is isosceles, if:bisector of $\angle BAC$ is perpendicular to base $BC.$
✓
Answer
In $\triangle \mathrm{ABC}$, the bisector of $\angle \mathrm{BAC}$ is perpendicular to the base $\mathrm{BC}$. We have to prove that the $\triangle A B C$ is isosceles.
In triangles $\mathrm{ADB}$ and $\mathrm{ADC}$,
$\angle \mathrm{BAD}=\angle \mathrm{CAD} \ldots . . .(\mathrm{AD}$ is bisector of $\angle \mathrm{BAC})$
$ \mathrm{AD}=\mathrm{AD} \ldots \ldots . .($Common$)$
$ \angle \mathrm{ADB}=\angle \mathrm{ADC} \ldots . . .($ Each equal to $90^{\circ})$
$ \Rightarrow \triangle \mathrm{ADB} \cong \triangle \mathrm{ADC} \ldots . . .($ by $\text{ASA}$ congruence criterion$)$
$ \Rightarrow \mathrm{AB}=\mathrm{AC} .......($cpct$)$
Hence, $\triangle \mathrm{ABC}$ is isosceles.
In triangles $\mathrm{ADB}$ and $\mathrm{ADC}$,
$\angle \mathrm{BAD}=\angle \mathrm{CAD} \ldots . . .(\mathrm{AD}$ is bisector of $\angle \mathrm{BAC})$
$ \mathrm{AD}=\mathrm{AD} \ldots \ldots . .($Common$)$
$ \angle \mathrm{ADB}=\angle \mathrm{ADC} \ldots . . .($ Each equal to $90^{\circ})$
$ \Rightarrow \triangle \mathrm{ADB} \cong \triangle \mathrm{ADC} \ldots . . .($ by $\text{ASA}$ congruence criterion$)$
$ \Rightarrow \mathrm{AB}=\mathrm{AC} .......($cpct$)$
Hence, $\triangle \mathrm{ABC}$ is isosceles.
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