Let $\angle BAC=\angle BDC=x \implies \angle ABD=\angle ACD =3x $, Let $O $ be the center of the circle, $ \angle AOD=6x $ We know, $$\angle EBD=120^{\circ}-x \implies \boxed{x=30^{\circ}} \implies \angle AOD=180^{\circ} \implies AO \in AD \text { and } \angle ABD=90^{\circ} $$Hence, $AD $ is the diameter of the circle

$\angle BAC= \angle BDC = x ; Also, \angle ABD = 3x$ $ In $$\Delta$ DBE by exterior angle theorem we get $x+60^{\circ} = 3x$ So $x=30^{\circ} \Rightarrow 3x= 90^{\circ} $ AD is a diameter