Find Angle ∢CDB in an Isosceles Right Triangle with Median

To solve this problem, we need to analyze the properties of the isosceles right triangle $\triangle ABC$:

• The triangle is isosceles and right ($\angle BAC = 90^\circ$), meaning $\angle ABC = \angle ACB = 45^\circ$.
• $BD$ is the median, dividing the hypotenuse $AC$ into two equal parts ($AD = DC$).

Now, we consider triangle $\triangle CDB$:

• Since $D$ is the midpoint of $AC$, $AD = DC$ and $BD$ is the median, which also acts as the altitude because the triangle is isosceles and right.
• Because $BD$ is an altitude in right triangle $\triangle ABC$, triangle $\triangle CDB$ is itself an isosceles right triangle, with $\angle DBC = \angle DCB$.

Using the properties of triangle $\triangle CDB$:

• The angle sum property of a triangle states that the sum of the angles in a triangle is $180^\circ$.
• $\angle CDB$ in triangle $\triangle CDB$ must therefore be $90^\circ$, as $\angle DBC = \angle DCB$ are both $45^\circ$.

Therefore, the size of $\angle CDB$ is $90^\circ$.