a) Xét $∆ABD$ và $∆ACD$ có:
$\widehat{BAD}=\widehat{CAD}=\dfrac12\widehat{BAC}\quad (gt)$
$\widehat{B}=\widehat{C}\quad (gt)$
$\Rightarrow 180^\circ -(\widehat{BAD} +\widehat{B})= 180^\circ -(\widehat{CAD} +\widehat{C})$
$\Rightarrow \widehat{ADB}=\widehat{ADC}$
$AD:$ cạnh chung
Do đó $∆BAD=∆CAD\, (g.c.g)$
b) Ta có: $∆BAD=∆CAD$ (câu a)
$\Rightarrow AB = AC$ (hai cạnh tương ứng)