Calculate Angle CDB in an Isosceles Right Triangle with Median

To solve this problem, we must understand the properties of an isosceles right triangle. In $\triangle ABC$, since it is an isosceles right triangle: $\angle ACB = \angle CAB = 45^\circ$ and $\angle ABC = 90^\circ$.

The median BD will divide the triangle into two smaller triangles, ABD and CBD. In $\triangle ABC$, since BD is a median from vertex B to side AC, and because AC is the hypotenuse of this right triangle, BD is equal in length to segments AD and DC. By the definition of the median and symmetry of the isosceles triangle, angles $\angle ADB$ and $\angle CDB$ are both right angles. Therefore, $\angle CDB$ is $90^\circ$.

Consequently, the measure of angle $\angle CDB$ is clearly $90^\circ$.

Therefore, the solution to the problem is $\angle CDB = 90^\circ$.