Calculate Angle ADB in an Isosceles Right Triangle with Median

The solution to find $\angle ADB$ can be determined by understanding the properties of the isosceles right triangle and the median.

Given triangle $\triangle ABC$ is an isosceles right triangle with $\angle ABC = 90^\circ$ and $\angle BAC = \angle ACB = 45^\circ$, BD, being the median from B to the hypotenuse AC, causes A and C to be equally distant from D. Consequently, triangle ABD is congruent to triangle BDC.

In this configuration, triangle ABD forms half of the isosceles right triangle. Due to symmetry and the properties of medians in such triangles, we have two equal angles at ADB and BDC.

Since $\triangle ABD$ and $\triangle BDC$ are congruent under isosceles conditions, we find that angle $\angle ADB$, like $\angle ABC$, must also be $\angle ADB = 90^\circ$.

Therefore, the solution to $\angle ADB$ is $\boxed{90}$ degrees.