The opposite angles of a cyclic quadrilateral are x and 3x. Find the angles.

In a cyclic quadrilateral, the sum of opposite angles is $180^\circ$.
Given $\angle WXY = x$ and $\angle WZY = 3x$.

The value of $x$ is $\boxed{\text{45}^\circ}$.

From part (b), $x = 45^\circ$.
So, $\angle WZY = 3x = 3(45^\circ) = 135^\circ$.

Since O is the center of the circle, OW, OY, and OZ are radii.
In $\triangle OYZ$, $OY = OZ$ (radii of the same circle). Therefore, $\triangle OYZ$ is an isosceles triangle, and its base angles are equal:

In $\triangle OWZ$, $OW = OZ$ (radii of the same circle). Therefore, $\triangle OWZ$ is an isosceles triangle, and its base angles are equal:

The angle $\angle WZY$ of the cyclic quadrilateral is the sum of $\angle WZO$ and $\angle OZY$.

We know $\angle WZY = 135^\circ$.
So, $\angle WZO + \angle OZY = 135^\circ$.
Substitute $\angle WZO$ with $\angle OWZ$ (since $\angle WZO = \angle OZW = \angle OWZ$) and $\angle OZY$ with $\angle OYZ$ (since $\angle OZY = \angle OYZ$).

Hence, it is proven that $\boxed{\angle OWZ + \angle OYZ = 135^\circ}$.