In trapezoid $A B C D, A B$ is parallel to $D C$ and $\angle D A F=90^{\circ}$. Point $E$ is on $D C$ so that $E B=B C=C E$. Point $F$ is on $A B$ so that $D F$ is parallel to $E B$. In degrees, what is the measure of $\angle F D A$ ?

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Since $E B=B C=C E, \triangle B C E$ is equilateral so $\angle B E C=60^{\circ}$.

Since $D F$ is parallel to $E B, \angle F D C=\angle B E C=60^{\circ}$.

Since $A B$ is parallel to $D C$ and $A D$ is perpendicular to $A B, A D$ is also perpendicular to $D C$, so $\angle A D C=90^{\circ}$.

Therefore, $\angle F D A=\angle A D C-\angle F D C=90^{\circ}-60^{\circ}=30^{\circ}$.

ANSWER: $30^{\circ}$