Take point $G$ on the opposite ray of ray $DC$ such that $GD = BE$. Then two triangles $ABE$, $ADG$ are congruent, implies that $AE = AG$ and $\angle EAB = \angle DAG = 20^{\circ}$. Thus
$$\angle FAG = \angle FAD + \angle DAG = \angle EAB + \angle DAG = 45^{\circ}.$$
Thus two triangles $AEF$, $AGF$ have common side $AF$ and $AE = AG$ and $\angle FAG = \angle FAE = 45^{\circ}$, thus $\triangle AEF \cong \triangle AGF$, then $AH = AD = 1$, where $AH$ is the altitude with respect to vertex $A$ in triangle $AEF$.

$\square$