About pentagon $ABCDE$ is known that angle $A$ and angle $C$ are right and that the sides $| AB | = 4$, $| BC | = 5$, $| CD | = 10$, $| DE | = 6$. Furthermore, the point $C'$ that appears by mirroring $C$ in the line $BD$, lies on the line segment $AE$. Find angle $E$.
1. **Construct the Pentagon and Reflect Point \( C \)**:
   - Given: \( \angle A = 90^\circ \) and \( \angle C = 90^\circ \).
   - Sides: \( AB = 4 \), \( BC = 5 \), \( CD = 10 \), \( DE = 6 \).
   - Reflect point \( C \) across line \( BD \) to get point \( C' \). Since \( C' \) lies on line segment \( AE \), we need to find the coordinates and properties of \( C' \).

2. **Determine Coordinates and Distances**:
   - Let \( B \) be at the origin \((0,0)\), \( A \) at \((4,0)\), and \( C \) at \((4,5)\) since \( \angle A = 90^\circ \).
   - \( D \) is at \((4, 5 + 10) = (4, 15)\) because \( CD = 10 \).

3. **Find the Perpendicular from \( C \) to \( BD \)**:
   - The line \( BD \) has a slope of \( \frac{15-0}{4-0} = \frac{15}{4} \).
   - The equation of line \( BD \) is \( y = \frac{15}{4}x \).
   - The perpendicular from \( C \) to \( BD \) will have a slope of \( -\frac{4}{15} \).
   - The equation of the perpendicular line from \( C \) is \( y - 5 = -\frac{4}{15}(x - 4) \).

4. **Find Intersection Point \( O \)**:
   - Solve the system of equations:
     \[
     y = \frac{15}{4}x \quad \text{and} \quad y - 5 = -\frac{4}{15}(x - 4)
     \]
   - Substituting \( y = \frac{15}{4}x \) into the second equation:
     \[
     \frac{15}{4}x - 5 = -\frac{4}{15}(x - 4)
     \]
     \[
     \frac{15}{4}x - 5 = -\frac{4}{15}x + \frac{16}{15}
     \]
     \[
     \frac{15}{4}x + \frac{4}{15}x = 5 + \frac{16}{15}
     \]
     \[
     \frac{225x + 16x}{60} = \frac{75 + 16}{15}
     \]
     \[
     \frac{241x}{60} = \frac{91}{15}
     \]
     \[
     241x = 364
     \]
     \[
     x = \frac{364}{241} = \frac{4}{3}
     \]
   - Substitute \( x = \frac{4}{3} \) back into \( y = \frac{15}{4}x \):
     \[
     y = \frac{15}{4} \cdot \frac{4}{3} = 5
     \]
   - So, \( O \) is at \( \left(\frac{4}{3}, 5\right) \).

5. **Calculate Distances \( BO \) and \( OD \)**:
   - \( BO = \sqrt{\left(\frac{4}{3}\right)^2 + 5^2} = \sqrt{\frac{16}{9} + 25} = \sqrt{\frac{16 + 225}{9}} = \sqrt{\frac{241}{9}} = \frac{\sqrt{241}}{3} \).
   - \( OD = \sqrt{\left(4 - \frac{4}{3}\right)^2 + (15 - 5)^2} = \sqrt{\left(\frac{8}{3}\right)^2 + 10^2} = \sqrt{\frac{64}{9} + 100} = \sqrt{\frac{64 + 900}{9}} = \sqrt{\frac{964}{9}} = \frac{\sqrt{964}}{3} \).

6. **Use Ptolemy's Theorem**:
   - For cyclic quadrilateral \( AC'OB \):
     \[
     AC' \cdot BO + AB \cdot OC' = AO \cdot BC'
     \]
   - Given \( AC' = 3 \), \( AB = 4 \), \( OC' = 2\sqrt{5} \), \( BO = \sqrt{5} \):
     \[
     3 \cdot \sqrt{5} + 4 \cdot 2\sqrt{5} = AO \cdot 5
     \]
     \[
     3\sqrt{5} + 8\sqrt{5} = 5AO
     \]
     \[
     11\sqrt{5} = 5AO
     \]
     \[
     AO = \frac{11\sqrt{5}}{5}
     \]

7. **Calculate \( \angle E \)**:
   - Using cosine rule in \( \triangle DEC' \):
     \[
     \cos \theta = \frac{4}{5}
     \]
   - Since \( \triangle DEC' \) is a \( 6-8-10 \) triangle, it is a right triangle.
   - Therefore, \( \angle E = 90^\circ \).

The final answer is \( \boxed{90^\circ} \).