Let $ABC$ be an isosceles right triangle with $\angle A=90^o$. Point $D$ is the midpoint of the side $[AC]$, and point $E \in [AC]$ is so that $EC = 2AE$. Calculate $\angle AEB + \angle ADB$ .
1. **Setup the problem using Cartesian coordinates:**
   - Place point \( B \) at the origin \((0,0)\).
   - Place point \( C \) at \((c,0)\).
   - Since \( \triangle ABC \) is an isosceles right triangle with \( \angle A = 90^\circ \), point \( A \) is at \((c/2, c/2)\).

2. **Determine the coordinates of points \( D \) and \( E \):**
   - Point \( D \) is the midpoint of \( AC \), so its coordinates are:
     \[
     D = \left( \frac{c/2 + c}{2}, \frac{c/2 + 0}{2} \right) = \left( \frac{3c}{4}, \frac{c}{4} \right)
     \]
   - Point \( E \) is such that \( EC = 2AE \). Let \( E = (x, y) \). Since \( E \in [AC] \), it lies on the line segment joining \( A \) and \( C \). The coordinates of \( E \) can be parameterized as:
     \[
     E = \left( \frac{c}{2} + t \cdot \frac{c}{2}, \frac{c}{2} - t \cdot \frac{c}{2} \right) = \left( \frac{c}{2}(1 + t), \frac{c}{2}(1 - t) \right)
     \]
     Given \( EC = 2AE \), we solve for \( t \):
     \[
     EC = \sqrt{\left( c - \frac{c}{2}(1 + t) \right)^2 + \left( 0 - \frac{c}{2}(1 - t) \right)^2} = \sqrt{\left( \frac{c}{2}(1 - t) \right)^2 + \left( \frac{c}{2}(1 - t) \right)^2} = \frac{c}{2} \sqrt{2(1 - t)^2}
     \]
     \[
     AE = \sqrt{\left( \frac{c}{2}(1 + t) - \frac{c}{2} \right)^2 + \left( \frac{c}{2}(1 - t) - \frac{c}{2} \right)^2} = \sqrt{\left( \frac{c}{2}t \right)^2 + \left( -\frac{c}{2}t \right)^2} = \frac{c}{2} \sqrt{2t^2}
     \]
     \[
     EC = 2AE \implies \frac{c}{2} \sqrt{2(1 - t)^2} = 2 \cdot \frac{c}{2} \sqrt{2t^2} \implies \sqrt{2(1 - t)^2} = 2 \sqrt{2t^2} \implies 1 - t = 2t \implies 1 = 3t \implies t = \frac{1}{3}
     \]
     Therefore, the coordinates of \( E \) are:
     \[
     E = \left( \frac{c}{2}(1 + \frac{1}{3}), \frac{c}{2}(1 - \frac{1}{3}) \right) = \left( \frac{2c}{3}, \frac{c}{3} \right)
     \]

3. **Calculate the angles \( \angle AEB \) and \( \angle ADB \):**
   - The slope of \( BE \) is:
     \[
     \text{slope of } BE = \frac{\frac{c}{3} - 0}{\frac{2c}{3} - 0} = \frac{1}{2}
     \]
     Thus, the angle \( \angle AEB \) is:
     \[
     \angle AEB = \tan^{-1} \left( \frac{1}{2} \right)
     \]
   - The slope of \( BD \) is:
     \[
     \text{slope of } BD = \frac{\frac{c}{4} - 0}{\frac{3c}{4} - 0} = \frac{1}{3}
     \]
     Thus, the angle \( \angle ADB \) is:
     \[
     \angle ADB = \tan^{-1} \left( \frac{1}{3} \right)
     \]

4. **Sum the angles \( \angle AEB \) and \( \angle ADB \):**
   - Since \( \angle AEB \) and \( \angle ADB \) are both angles formed with the x-axis, their sum is:
     \[
     \angle AEB + \angle ADB = \tan^{-1} \left( \frac{1}{2} \right) + \tan^{-1} \left( \frac{1}{3} \right)
     \]
   - Using the tangent addition formula:
     \[
     \tan^{-1} \left( \frac{1}{2} \right) + \tan^{-1} \left( \frac{1}{3} \right) = \tan^{-1} \left( \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \cdot \frac{1}{3}} \right) = \tan^{-1} \left( \frac{\frac{3 + 2}{6}}{1 - \frac{1}{6}} \right) = \tan^{-1} \left( \frac{\frac{5}{6}}{\frac{5}{6}} \right) = \tan^{-1}(1) = 45^\circ
     \]

5. **Final calculation:**
   - Since \( \angle AEB + \angle ADB = 45^\circ \) and the right angle \( \angle ACB = 90^\circ \), the total angle is:
     \[
     \angle AEB + \angle ADB + \angle ACB = 45^\circ + 90^\circ = 135^\circ
     \]

The final answer is \( \boxed{135^\circ} \)