Question
In $\triangle ABC,\text{AB} = \text{AC}.$ Show that the altitude $\text{AD}$ is median also.
✓
Answer
In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{ADC}$,
$\mathrm{AB}=\mathrm{AC} \quad.... ($ Since is ani isosceles triangle$)$
$\mathrm{AD}=\mathrm{AD} \quad..... ($ common side $)$
$\angle ADB=\angle ADC \quad \ldots .\left(\right.$ Since $\text{AD}$ is the altitude so each is $\left.90^{\circ}\right)$
$\Rightarrow \triangle ADB \cong \triangle ADC \quad..... (\text{RHS}$ congruence criterion $)$
$\mathrm{BD}=\mathrm{DC} \quad.... ($cpct$)$
$\Rightarrow \text{AD}$ is the median.
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