Prove CD = CE, given that line passing through C intersects with angle bisectors at D and E Let $PA$ and $PB$ be the tangents to a circle centered at $O$ and $C$ is a point on the minor arc $AB$. The line passing through the point $C$ and perpendicular to the line $PC$ intersects with the internal angle bisectors of $\angle{AOC}$ and $\angle{BOC}$ at the points $D$ and $E$, respectively. Show that $CD=CE$

I am trying to prove that the area of $\triangle{PCD}$ is equal to the area of $\triangle{PCE}$, so that $CD/CE=1$. 
But, I'm having problem proving $\angle{PDC}=\angle{PEC}$ or $\angle{DPC}=\angle{EPC}$. Please help!
 A: 
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
$$CD=CE$$
A: This is a difficult question. I can only finish half of it and the last paragraph is a suggestion in arriving at the goal.

Since all the green marked angles are equal, we can say that BOCQ is cyclic; and hence $\angle OCQ = 90^0$.
The angle bisectors create kites. From them, we get (1) all red marked angles are equal; (2) AD = DC; (3) CE = EB.
The job is done if we can show $\triangle PDA \cong \triangle PEB$. To this end, we have (1) $\angle PAD = \angle PBE$; and (2) PA = PB already. It remains to show that $\angle PDA = \angle PEB$. Or equivalently, if we can show that PEDR (where R is the intersection point of BE and AD produced) is cyclic. Good luck. 
