How to prove the two segments are of equal length? I met a primary geoemtry problem below, which seems too difficult for me to find a proof: 
As in the figure, $P$ is a point outside circle $O$; 
$PA$ and $PB$ are tangent lines of circle $O$; $C$ is a point on the inferior arc ${\stackrel{{\mbox{$\Large{\frown}$}}}{AB}}$ distinct from $A$ or $B$;  $OD$ and $OE$ bisect $\angle AOC$ and $\angle BOC$ respectively; $PC\bot DE$. 
Prove: $\overline{CD}=\overline{CE}$;
Answers/hints in primary mathematics are expected.

 A: Three different approaches:
(1)
Problem solved by using sine theorem several times.
Applying sine theorm to $\Delta OCD$ and $\Delta OCE$:
$\dfrac{\overline{CD}}{\sin{\alpha}}=\dfrac{\overline{OC}}{\sin{\angle{ODC}}}$  and 
$\dfrac{\overline{CE}}{\sin{\beta}}=\dfrac{\overline{OC}}{\sin{\angle{OEC}}}$,
Then:
$\overline{CD}=\overline{CE}\Leftrightarrow {}\dfrac{\overline{OC}\sin{\alpha}}{\sin{\angle{ODC}}} ={}\dfrac{\overline{OC}\sin{\beta}}{\sin{\angle{OEC}}}\Leftrightarrow \dfrac{\sin{\alpha}}{\sin{\angle{ODC}}} ={}\dfrac{\sin{\beta}}{\sin{\angle{OEC}}}$
Applying sine theorm to $\Delta APC$ and $\Delta BPC$:
$$\frac{\overline{PC}}{\sin{\alpha}}=\frac{\overline{PA}}{\sin\angle{ACP}},\quad \frac{\overline{PC}}{\sin{\beta}}=\frac{\overline{PB}}{\sin\angle{BCP}} \Rightarrow \dfrac{\sin{\alpha}}{\sin{\angle{ACP}}} ={}\dfrac{\sin{\beta}}{\sin{\angle{BCP}}}$$
So we only need to prove:
$$\frac{\sin{\angle{ODC}}}{\sin\angle{OEC}}=\frac{\sin\angle{ACP}}{\sin\angle{BCP}}$$
Since $\angle{ACP}=\left(\dfrac{\pi}{2}-\angle{ODC}\right)+\dfrac{\pi}{2},\angle{BCP}=\left(\dfrac{\pi}{2}-\angle{OEC}\right)+\dfrac{\pi}{2}$, this proves the problem.

(2) using cosine theorem

Since $\overline{CD}=\overline{CE}\Leftrightarrow \overline{AD}=\overline{BE}$, problem is converted into the new one. 
Easy to prove $\overline{PD}^2-\overline{AD}^2=\overline{PC}^2=\overline{PE}^2-\overline{BE}^2$ and let $\theta=\angle{PAD}=\angle{OCD}-\dfrac{\pi}{2}=\dfrac{\pi}{2}-\angle{OCE}=\angle{PBE}$, then apply cosine theorem to $\Delta PAD$ and $\Delta PBE$ to compare the expressions of $\cos\theta$ we can obtain $\overline{AD}=\overline{BE}$.
Note the case when $C$  exactly bisects arc $AB$ is trivial.
(3) 
Further auxiliary conductors:

(4)
new solution:

