Suppose the angle bisector of $\angle{ABD}$ intersect $AC$ at $E$, we have $CE=CB$ So $AC>BC\rightarrow \angle{BAC}<\angle{ABC}\rightarrow \angle{BAC}<90^{\circ}$. Let $AC=b,AB=c,BC=a$, we get $\frac{AE}{ED}=\frac{AB}{BD}$. So $\frac{b-a}{a-\frac{b}{2}}=\frac{c}{\sqrt{\frac{2a^2+2c^2-b^2}{4}}}\rightarrow (2a^2-b^2)(a-b-c)(a-b+c)=0$. We have $b^2=2a^2$. Note that $cos(\angle{BAC}) =\frac{b^2+c^2-a^2}{2bc}=\frac{a^2+c^2}{2\sqrt{2}ac}\geq \frac{1}{\sqrt{2}}$. So $\angle{BAC}\leq 45^{\circ}$. Equality holds at $a=c=\frac{b}{\sqrt{2}}$.

ThE-dArK-lOrD wrote: We have $b^2=2a^2$. This equality we can get by follow way: If $X$ is point of intersection of bisectors of $\angle ACB,\angle ABD$, then $90^\circ = \angle BAC + (\angle ABD + \angle ACB)/2 = \angle CBD + (\angle ABD + \angle ACB)/2 \Rightarrow \angle BAC = \angle CBD$, so $\Delta ACB \sim \Delta BCD$ and $b^2=2a^2$.