Question
Prove that a $\triangle ABC$ is isosceles, if: altitude $AD$ bisects angles $BAC.$
✓
Answer
In $\triangle A B C$, let the altitude $A D$ bisects $\angle B A C$. Then we have to prove that the $\triangle A B C$ is isosceles.
In triangles $\mathrm{ADB}$ and $\mathrm{ADC}$,
$\angle B A D=\angle C A D \ldots(A D$ is bisector of $\angle B A C)$
$A D=A D \ldots ($common$)$
$\angle A D B=\angle A D C \ldots .($ Each equal to $90^{\circ})$
$\Rightarrow \triangle A D B \cong \triangle A D C \ldots ($by $\text{ASA}$ congruence criterion$)$
$\Rightarrow A B=A C \ldots ($cpct$)$
Hence, $\triangle \mathrm{ABC}$ is an isosceles.
In triangles $\mathrm{ADB}$ and $\mathrm{ADC}$,
$\angle B A D=\angle C A D \ldots(A D$ is bisector of $\angle B A C)$
$A D=A D \ldots ($common$)$
$\angle A D B=\angle A D C \ldots .($ Each equal to $90^{\circ})$
$\Rightarrow \triangle A D B \cong \triangle A D C \ldots ($by $\text{ASA}$ congruence criterion$)$
$\Rightarrow A B=A C \ldots ($cpct$)$
Hence, $\triangle \mathrm{ABC}$ is an isosceles.
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