Xét $ΔABD$ và $ΔACD$ có:
$\widehat{ABD} = \widehat{ACD} \quad (gt)$
$\widehat{BAD} = \widehat{CAD} = \dfrac12\widehat{BAC}\quad (gt)$
$\to 180^\circ - (\widehat{ABD} + \widehat{BAD}) = 180^\circ - (\widehat{ACD} + \widehat{CAD})$
$\to \widehat{ADB} = \widehat{ADC}$
$AD:$ cạnh chung
Do đó $ΔABD = ΔACD\, (g.c.g)$
$\to AB = AC\quad$ (hai cạnh tương ứng)