claim: $CD^2=CE \cdot CA$. By the Law of Sines :$\frac{EC}{\sin60^{\circ}}=\frac{BC}{\sin40^{\circ}}$, $\frac{CD}{\sin80^{\circ}}=\frac{BC}{\sin30^{\circ}}.$ Since $AC=\frac{2BC}{\sin10^{\circ}}$ $CD^2=CE \cdot CA \Leftrightarrow 4\sin^2 80^{\circ}=\frac{\sqrt3}{4\sin10^{\circ} \sin40^{\circ}}$. $\Leftrightarrow 4\sin20^{\circ}\cos10^{\circ}\sin40^{\circ}=\frac{\sqrt3}{2}=\cos30^{\circ}$. $\Leftrightarrow -2(\cos60^{\circ}-\cos20^{\circ})\cos10^{\circ} = \cos20^{\circ}\cos10^{\circ}-\sin20^{\circ}\sin10^{\circ}$. $\Leftrightarrow \cos10^{\circ} = \cos20^{\circ}\cos10^{\circ}+\sin20^{\circ}\sin10^{\circ}$, which is trivial. $CD^2=CE \cdot CA \Rightarrow \triangle CDE \sim \triangle CAD$. This implies that $\angle CDE =\boxed {20^{\circ}}$.

Ceva trig. is enough to solve this problem, how did this become a P8?

Haha! For a second I thought someone had posted here one of my videos... https://youtu.be/8Mt-i6rEiHI