Find $x$ .

Beautiful!! Let $AF \cap \odot ABC = D$ and $I$ be the incenter of $\Delta BCD.$ Clearly, we have $\Delta AEF \cong \Delta BIC \Rightarrow BI = BF \Longrightarrow \angle BIF = 80^{\circ} \therefore x= 30^{\circ}$

Let the side length of $\triangle ABC$ be 1 and $AE=BF=a$ . By sine rule, we have:

$\begin{aligned} \frac {BF}{\sin \angle BAF} & = \frac {AB}{\sin \angle AFB} \\ \frac a{\sin 20^\circ} & = \frac 1{\color{#3D99F6}\sin 100^\circ} & \small \color{#3D99F6} \text{Note that }\sin (180^\circ - \theta) = \sin \theta \\ \implies a & = \frac {\color{#D61F06}\sin 20^\circ}{\color{#3D99F6}\sin 80^\circ} & \small \color{#3D99F6} \text{And }\sin (90^\circ - \theta) = \cos \theta \\ & = \frac {\color{#D61F06}2\sin 10^\circ\cos 10^\circ}{\color{#3D99F6}\cos 10^\circ} & \small \color{#D61F06} \text{Also }\sin (2 \theta) = 2\sin \theta \cos \theta \\ \implies a & = 2\sin 10^\circ \end{aligned}$

By sine rule again,

$\begin{aligned} \frac {\sin \angle AEB}{AB} & = \frac {\sin \angle ABE}{AE} \\ \frac {\color{#3D99F6}\sin (180^\circ - x)}1 & = \frac {\sin (x-20^\circ)}{\color{#D61F06}a} & \small \color{#3D99F6} \text{Note that }\sin (180^\circ - \theta) = \sin \theta \\ \frac {\color{#3D99F6}\sin x}1 & = \frac {\sin (x-20^\circ)}{\color{#D61F06}2\sin 10^\circ} & \small \color{#D61F06} a = 2\sin 10^\circ \\ \frac {\sin x}{\sin (x-20^\circ)} & = \frac {\frac 12}{\sin 10^\circ} \\ & = \frac {\sin 30^\circ}{\sin (30^\circ - 20^\circ)} \\ \implies x & = \boxed{30}^\circ \end{aligned}$