Let $(O,R)$ be the circumcircle of $\triangle ABC,$ $M$ the midpoint of $BC$ and $P$ the foot of the A-altitude. $\angle ADB=\angle PAD=45^{\circ}$ $\Longrightarrow$ $\angle PAO=90^{\circ}$ $\Longrightarrow$ $AO \parallel BC,$ i.e. $OM=AP.$ On the other hand, we have: $AD^2=AB \cdot AC - BD \cdot CD=AB \cdot AC-AD^2$ $\Longrightarrow$ $2 \cdot AD^2= AB \cdot AC$ $\Longrightarrow$ $AD^2= 2 \cdot AP^2= R \cdot AP$ $\Longrightarrow$ $2 \cdot OM=R$ $\Longrightarrow$ $\angle OCM=30^{\circ}$ Therefore, $\angle A=60^{\circ} \ , \angle B=105^{\circ}$ and $\angle C=15^{\circ}.$