Use the angle sum property of a triangle.

In any triangle ABC, the sum of angles is $A+B+C = 180^\circ$.
Therefore, $B = 180^\circ - (A+C)$.
Using the cosine identity $\cos(180^\circ - \theta) = -\cos \theta$, we have:

$$\cos B = \cos(180^\circ - (A+C)) = -\cos(A+C)$$

The given equation is $\frac{\sin A + \cos A}{\cos B} = \sqrt{2}$.
Substitute $\cos B = -\cos(A+C)$:

$$\frac{\sin A + \cos A}{-\cos(A+C)} = \sqrt{2}$$ $$\sin A + \cos A = -\sqrt{2} \cos(A+C)$$

Using the cosine addition formula $\cos(X+Y) = \cos X \cos Y - \sin X \sin Y$:

$$\sin A + \cos A = -\sqrt{2} (\cos A \cos C - \sin A \sin C)$$ $$\sin A + \cos A = -\sqrt{2} \cos A \cos C + \sqrt{2} \sin A \sin C$$

Rearrange the terms to group $\sin A$ and $\cos A$:

$$\sin A - \sqrt{2} \sin A \sin C + \cos A + \sqrt{2} \cos A \cos C = 0$$

Factor out $\sin A$ and $\cos A$:

$$\sin A (1 - \sqrt{2} \sin C) + \cos A (1 + \sqrt{2} \cos C) = 0$$

For the equation $\sin A (1 - \sqrt{2} \sin C) + \cos A (1 + \sqrt{2} \cos C) = 0$ to hold true for any angle $A$ in a triangle, the coefficients of $\sin A$ and $\cos A$ must both be zero, as $\sin A$ and $\cos A$ are linearly independent.
Thus, we must have:

$$1 - \sqrt{2} \sin C = 0 \quad \text{(Equation 1)}$$ $$1 + \sqrt{2} \cos C = 0 \quad \text{(Equation 2)}$$

From Equation 1:

$$\sqrt{2} \sin C = 1 \implies \sin C = \frac{1}{\sqrt{2}}$$

From Equation 2:

$$\sqrt{2} \cos C = -1 \implies \cos C = -\frac{1}{\sqrt{2}}$$

We need an angle $C$ such that $\sin C = \frac{1}{\sqrt{2}}$ and $\cos C = -\frac{1}{\sqrt{2}}$.
Since $\sin C$ is positive and $\cos C$ is negative, angle $C$ must be in the second quadrant.
The reference angle for which $\sin \theta = \frac{1}{\sqrt{2}}$ and $\cos \theta = \frac{1}{\sqrt{2}}$ is $45^\circ$.
In the second quadrant, the angle is $180^\circ - 45^\circ = 135^\circ$.
Therefore, $C = 135^\circ$.
This value is valid for an angle in a triangle ($0^\circ < C < 180^\circ$).

Thus, we have proven that $C = 135^\circ$.

The final answer is \boxed{\text{C = 135^\circ}}