Let $\alpha$ and $\beta$ be the bisected angles as marked in the diagram, which shows $2\alpha + \theta = 2\beta - \theta$, or
$$\beta - \alpha = \theta$$
Apply the sine rule to the triangles APO and CPO,
$$\frac{r}{PO} = \frac{\sin\angle CPO}{\sin(90+x)}
=\frac{\sin\angle APO}{\sin 90}\tag{1}$$
Recognize $\angle CPO = 90-\theta-x$ and $\angle APO = 90-\alpha-\beta$ to reexpress (1) as
$$\cos(\alpha+\beta)\cos x = \cos(x+\beta-\alpha)$$
Rewrite both sides,
$$\begin{split}
\text{lhs}&=\frac 12\left[ \cos(x+\alpha+\beta)+\cos(x-\alpha-\beta)\right]\\ \text{rhs}&=\frac 12 \left[ \cos(x+\alpha+\beta) + \cos(x-\alpha+\beta)\right]
+ \sin(x+\beta)\sin\alpha
\end{split}$$
After canceling the common terms and collapsing $\cos(x-\alpha-\beta)-\cos(x-\alpha+\beta)$, we get,
$$\sin\beta\sin(x-\alpha)=\sin\alpha\sin(x+\beta) \tag{2}$$
Now, apply the sine rule to the triangles OCD and OCE,
$$\frac{CD}{r}= \frac{\sin\alpha}{\sin (x-\alpha)},\>\>\>
\frac{CE}{r}= \frac{\sin\beta}{\sin (x+\beta)}$$
and take their ratio,
$$\frac{CD}{CE}=\frac{\sin\alpha\sin (x+\beta)}{\sin\beta\sin (x-\alpha)} $$
Use the result (2) to get
$$CD=CE$$