In triangle AB=AC and ad perpendicular to BC prove that D is the midpoint of BC

$\mathrm{In}\u2206\mathrm{ADB}\mathrm{and}\u2206\mathrm{ADC}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\angle \mathrm{ADB}=\angle \mathrm{ADC}\left[90\xb0\mathrm{each}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{AB}=\mathrm{AC}\left[\mathrm{Given}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{AD}=\mathrm{AD}\left[\mathrm{Common}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{So},\u2206\mathrm{ADB}\cong \u2206\mathrm{ADC}\left[\mathrm{RHS}\right]$

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