The Geometry Of Coffee Cup Rings

By symmetry, $\angle ADB = \angle ABD = 20^{\circ}$ . Then by the central angle theorem ,

$x = \angle AED = \dfrac{1}{2} \angle ADB = \boxed{10^{\circ}}$ .

Alternatively, we can note that $\angle ADE = 180^{\circ} - \angle ADB = 160^{\circ}$ , and that since $|AD| = |AE|$ we know that $\Delta ADE$ is an isosceles triangle with $\angle DAE = \angle AED = x$ . Thus

$2x + \angle ADE = 180^{\circ} \Longrightarrow 2x + 160^{\circ} = 180^{\circ} \Longrightarrow x = \boxed{10^{\circ}}$ .