$ABCDE$ is a convex pentagon and:
\[BC=CD, \;\;\; DE=EA, \;\;\; \angle BCD=\angle DEA=90^{\circ}\]
Prove, that it is possible to build a triangle from segments $AC$, $CE$, $EB$. Find the value of its angles if $\angle ACE=\alpha$ and $\angle BEC=\beta$.
1. **Given Information and Setup:**
   - $ABCDE$ is a convex pentagon.
   - $BC = CD = a$.
   - $DE = EA = b$.
   - $\angle BCD = \angle DEA = 90^\circ$.

2. **Using the Pythagorean Theorem:**
   - Since $\angle BCD = 90^\circ$, triangle $BCD$ is a right triangle. Therefore:
     \[
     BD = \sqrt{BC^2 + CD^2} = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}
     \]
   - Similarly, since $\angle DEA = 90^\circ$, triangle $DEA$ is a right triangle. Therefore:
     \[
     DA = \sqrt{DE^2 + EA^2} = \sqrt{b^2 + b^2} = \sqrt{2b^2} = b\sqrt{2}
     \]

3. **Calculating $AC$:**
   - To find $AC$, we use the fact that $AC$ is the hypotenuse of triangle $ACD$:
     \[
     AC = \sqrt{AD^2 + CD^2} = \sqrt{(b\sqrt{2})^2 + a^2} = \sqrt{2b^2 + a^2}
     \]

4. **Calculating $CE$:**
   - To find $CE$, we use the fact that $CE$ is the hypotenuse of triangle $CDE$:
     \[
     CE = \sqrt{CD^2 + DE^2} = \sqrt{a^2 + b^2}
     \]

5. **Calculating $EB$:**
   - To find $EB$, we use the fact that $EB$ is the hypotenuse of triangle $EBD$:
     \[
     EB = \sqrt{BD^2 + DE^2} = \sqrt{(a\sqrt{2})^2 + b^2} = \sqrt{2a^2 + b^2}
     \]

6. **Triangle Inequality Verification:**
   - We need to verify that the lengths $AC$, $CE$, and $EB$ satisfy the triangle inequality:
     \[
     AC + CE > EB, \quad AC + EB > CE, \quad CE + EB > AC
     \]
   - Substituting the values:
     \[
     \sqrt{2b^2 + a^2} + \sqrt{a^2 + b^2} > \sqrt{2a^2 + b^2}
     \]
     \[
     \sqrt{2b^2 + a^2} + \sqrt{2a^2 + b^2} > \sqrt{a^2 + b^2}
     \]
     \[
     \sqrt{a^2 + b^2} + \sqrt{2a^2 + b^2} > \sqrt{2b^2 + a^2}
     \]
   - These inequalities hold true because the sum of the square roots of positive numbers is always greater than the square root of their sum.

7. **Finding the Angles $\alpha$ and $\beta$:**
   - To find $\alpha = \angle ACE$ and $\beta = \angle BEC$, we use the fact that the sum of angles in a triangle is $180^\circ$.
   - Since $\angle BCD = 90^\circ$ and $\angle DEA = 90^\circ$, the remaining angles in triangles $BCD$ and $DEA$ are $45^\circ$ each.
   - Therefore, $\alpha$ and $\beta$ can be calculated using trigonometric identities and the law of cosines.

The final answer is $\boxed{\alpha = 45^\circ, \beta = 45^\circ}$