1. Define the angles:
Let $\angle ABE = \angle EBC = \theta$. Given that $\angle BEA = 45^\circ$, we can determine the other angles in the triangle.

2. **Determine $\angle BAC$:**
Since $\angle BEA = 45^\circ$ and $\angle ABE = \theta$, we have:
$$\angle BAC = 180^\circ - \angle ABE - \angle BEA = 180^\circ - \theta - 45^\circ = 135^\circ - \theta$$

3. **Determine $\angle BCA$:**
Using the fact that the sum of angles in a triangle is $180^\circ$, we have:
$$\angle BCA = 180^\circ - \angle BAC - \angle ABC = 180^\circ - (135^\circ - \theta) - \theta = 45^\circ - \theta$$

4. **Determine $\angle HAC$:**
Since $AH$ is an altitude, $\angle HAC$ is complementary to $\angle BCA$:
$$\angle HAC = 90^\circ - \angle BCA = 90^\circ - (45^\circ - \theta) = 45^\circ + \theta$$

5. Use the sine rule:
Since $\sin(135^\circ - \theta) = \sin(45^\circ + \theta)$, we have:
$$\frac{AB}{BC} = \frac{AH}{HC}$$

6. Use the angle bisector theorem:
Since $BE$ is an angle bisector, we have:
$$\frac{AE}{EC} = \frac{AB}{BC}$$

7. **Relate $HE$ to $AH$ and $HC$:**
From the above, we know that $HE$ is the angle bisector of $\angle AHC$. Since $\angle AHC = 90^\circ$, the angle bisector $HE$ divides it into two equal angles:
$$\angle EHC = \frac{1}{2} \times 90^\circ = 45^\circ$$

Thus, we have shown that $\angle EHC = 45^\circ$.

$\blacksquare$