Let $\angle{CDE}=\alpha$. Then $\angle{ADC}=(140^\circ - \alpha)$. Then \begin{align} \frac{\sin\alpha}{\sin(140^\circ - \alpha)}&=\frac{CE}{AE}\\ &=\frac{BC\cdot\sin 50^\circ/\sin\angle{BEC}}{AB\cdot\sin 30^\circ /\sin\angle{BEA}}&\text{(by the law of sines)}\\ &=\frac{BC\cdot\sin 50^\circ}{AB\cdot\sin 30^\circ }\\ &=\frac{\sin 20^\circ}{\sin 80^\circ}\cdot\frac{\sin 50^\circ}{\sin 30^\circ}&\text{(by the law of sines)}\\ &=\frac{2\sin 20^\circ \cos 40^\circ}{\cos 10^\circ} \end{align} Hence \begin{align*} \sin\alpha\cos 10^\circ &= 2\sin 20^\circ\cos 40^\circ\sin(40^\circ + \alpha)\\ \sin\alpha\cos 10^\circ &= \sin 20^\circ(\sin(80^\circ + \alpha) + \sin\alpha)\\ \sin\alpha &= 2\sin 10^\circ (\sin(80^\circ + \alpha) + \sin\alpha)\\ \sin\alpha &= 2\sin 10^\circ \cos(10^\circ - \alpha) + 2\sin 10^\circ\sin\alpha\\ \sin\alpha &= \sin(20^\circ - \alpha) + \sin\alpha + 2\sin 10^\circ\sin\alpha\\ 0 &= \sin(20^\circ - \alpha) - 2\sin 10^\circ\sin\alpha\\ 0 &= \sin 20^\circ\cos\alpha - \cos 20^\circ\sin\alpha - 2\sin 10^\circ\sin\alpha \end{align*} \begin{align*} \therefore\quad \tan\alpha &= \frac{\sin 20^\circ}{\cos 20^\circ - 2\sin 10^\circ}\\ &= \frac{\sin 20^\circ}{\sin 70^\circ - \sin 10^\circ - \sin 10^\circ}\\ &= \frac{\sin 20^\circ}{2\sin 30^\circ \cos 40^\circ - \sin 10^\circ}\\ &= \frac{\sin 20^\circ}{\sin 50^\circ - \sin 10^\circ}\\ &= \frac{\sin 20^\circ}{2\sin 20^\circ \cos 30^\circ}\\ &= \frac{\sqrt{3}}{3}\\ \end{align*} Therefore we conclude $\alpha = \boxed{30^\circ}$.