Spring Mathematical Tournament

Since $(ABD) \perp (ABC), (BCD)$, then $(ABC) \perp BC$ and hence $\angle ABC = \angle DBC = 90^\circ$. Since $\angle BDC = 45^\circ$, then $BC = BD = 2$. Let $\angle ADB = \alpha$. Then $AB = 2 \sin \alpha$, $AD = 2 \cos \alpha$ and $AC = 2\sqrt{\sin^2 \alpha + 1}$. So
$$S_{ABD} = \frac{AB \cdot AD}{2} = 2 \sin \alpha \cdot \cos \alpha,$$
$$S_{ACD} = \frac{AC \cdot AD}{2} = 2 \cos \alpha \sqrt{1 + \sin^2 \alpha},$$
$$S_{BCD} = \frac{BC \cdot BD}{2} = 2.$$
It follows by the condition of the problem that
$$8 = 4(\sin^2 \alpha \cos^2 \alpha + \cos^2 \alpha(1 + \sin^2 \alpha) + 1) = 4(\sin^2 \alpha - 2 \sin^4 \alpha + 2),$$
i.e. $\sin \alpha = \frac{\sqrt{2}}{2}$. Thus $\angle ADB = 45^\circ$.