Angles inside a triangle

Since $\triangle ABC$ is isosceles with $CA=CB$ $\implies \angle CAB = \angle CBA = \dfrac {180^\circ - 20^\circ}2 = 80^\circ$ . Therefore, $\angle BEA = 180^\circ - 80^\circ-50^\circ = 50^\circ$ implies that $\triangle ABE$ is isosceles with $AB=EB$ . And also $\angle EBD = 80^\circ - 60^\circ = 20^\circ$ implies that $\triangle BCD$ is isosceles with $BD=CD$ .

Let $\angle EDB = \theta$ and $AB=EB=1$ . Using sine rule on $\triangle ABD$ :

$\begin{aligned} \frac {BD}{\sin \angle BAD} & = \frac {AB}{\sin \color{#3D99F6} \angle ADB} & \small \color{#3D99F6} \text{Note that }\angle ADB = 180^\circ - 80^\circ - 60^\circ = 40^\circ \\ \frac {BD}{\sin 80^\circ} & = \frac 1{\sin \color{#3D99F6} 40^\circ} \\ \implies BD & = \frac { \color{#3D99F6} \sin 80^\circ}{\sin 40^\circ} & \small \color{#3D99F6} \text{Note that } \sin 80^\circ = 2 \sin 40^\circ \cos 40^\circ \\ & = 2\cos 40^\circ \end{aligned}$

Using sine rule on $\triangle BDE$ :

$\begin{aligned} \frac {\sin \angle EDB}{EB} & = \frac {\sin \color{#3D99F6} \angle DEB}{BD} & \small \color{#3D99F6} \text{Note that }\angle DEB = 180^\circ - 20^\circ - \theta = 160^\circ - \theta \\ \frac {\sin \theta}{1} & = \frac {\color{#3D99F6} \sin (160^\circ - \theta)}{2\cos 40^\circ} & \small \color{#3D99F6} \text{Note that }\sin (180^\circ - x) = \sin x \\ 2\cos 40^\circ \sin \theta & = \color{#3D99F6} \sin (20^\circ + \theta) \\ & = \sin 20^\circ \cos \theta + \cos 20^\circ \sin \theta \\ \frac {\sin \theta}{\cos \theta} & = \frac {\sin 20^\circ}{2\cos 40^\circ - \cos 20^\circ} \\ \tan \theta & = \frac {\sin (30^\circ - 10^\circ)}{2\cos (30^\circ + 10^\circ) - \cos (30^\circ - 10^\circ)} \\ & = \frac {\sin 30^\circ \cos 10^\circ - \cos 30^\circ \sin 10^\circ }{\cos 30^\circ \cos 10^\circ - 3 \sin 30^\circ \sin 10^\circ} \\ & = \frac {\frac 12 \cos 10^\circ - \frac {\sqrt 3}2 \sin 10^\circ }{\frac {\sqrt 3}2 \cos 10^\circ - \frac 32 \sin 10^\circ} \\ & = \frac 1{\sqrt 3} \\ \implies \theta & = \boxed{30^\circ} \end{aligned}$