We'll use the formula for the angle bisector (this can be derived using Stewart's theorem): $AD=\sqrt{AB.AC-BD.CD}$. On the other hand $AD=\sqrt{BD.CD}\implies AB.AC=2BD.CD\implies bc=2\left(\frac{ba}{b+c}\right)\left(\frac{ca}{b+c}\right)\implies (b+c)^2=2a^2$. Let $\angle BAC=2\alpha$ and then the side equality becomes equivalent to: $$\left( \sin(45^{\circ}-\alpha)+\sin(135^{\circ}-\alpha)\right)^2=2( \sin(2\alpha))^2$$$$\implies \left(\sin(45^{\circ}-\alpha)+\sin(45^{\circ}+\alpha)\right)^2=2(\sin(2\alpha))^2$$We know that $0^{\circ}<\alpha<45^{\circ}$. It's obvious that the function $f(x)=2\sin(2x)^2$ is strictly increasing in the interval $0<x<\frac{\pi}{4}$ and the function $g(x)=(\sin(x+\frac{\pi}{4})+\sin(x-\frac{\pi}{4})$ is strictly decreasing in the same interval. Therefore we only need to find one value of $x$ for which $f(x)=g(x)$ and we would have solved the problem. Now we have to notice that $\left(\sin(15^{\circ})+\sin(75^{\circ})\right)^2=\left(\frac{\sqrt{3}-1}{2\sqrt{2}}+\frac{\sqrt{3}+1}{2\sqrt{2}}\right)^2=\left(\frac{\sqrt{3}}{\sqrt{2}}\right)^2=2\left(\sin(60^{\circ})\right)^2$, so the angles of $\triangle ABC$ are $\angle A=60^{\circ}$, $\angle B=105^{\circ}$, $\angle C=15^{\circ}$.

Let $O, M$ - circumcenter and midpoint of the arc $BC$. By PoP we have $AD=DM$. So $OD\perp AM$, $\angle ODC=45^\circ$, $O$ - reflection of $M$ over $BC$. Hence $\angle A=60^\circ$. If $AD$ - diameter, then $AD\parallel BC$. And $$120^\circ-\angle C=\angle B=\angle ABD+\angle CBD=90^\circ +\angle C$$

parmenides51 wrote: In triangle $ABC$, the length of the angle bisector $AD$ is $\sqrt{BD \cdot CD}$. Find the angles of the triangle $ABC$, if $\angle ADB = 45^o$. $\color{blue}\boxed{\textbf{Answer: 60, 105, 15}}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$AD^2=BD.CD...(I)$$$$\angle BAD=\angle DAC=\alpha$$Let $T$ be a point in $BC$ (it can be in its prolongation) such that $\angle ATC=\alpha$, note that $T$ exists from $\alpha<90(\angle BAC<180)$ Like $\angle ATB=\angle BAD$ $$\Rightarrow AD^2=BD.TD$$By $(I):$ $$\Rightarrow BD.TD=BD.CD$$$$\Rightarrow TD=CD=a$$Like $\angle ATD=\angle DAC$ $$\Rightarrow AC^2=DC.TC=a.(2a)=2a^2$$$$\Rightarrow AC=a\sqrt{2}$$Like $\angle ADB=45$ and $\angle DAC=\alpha$ $$\Rightarrow \angle DCA=45-\alpha$$In $\triangle ACD:$ $$\Rightarrow \frac{S_{\alpha}}{S_{135}}=\frac{a}{a\sqrt{2}}$$$$\Rightarrow S_{\alpha}=\frac{1}{2}, \alpha \in [0,90]$$$$\Rightarrow \alpha=30$$$$\Rightarrow \boxed{\textbf{The angles of ABC are: 60, 105, 15}}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$