SAUDI ARABIAN MATHEMATICAL COMPETITIONS

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 $△AEF\cong △AGF$, then $AH=AD=1$, where $AH$ is the altitude with respect to vertex $A$ in triangle $AEF$.

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