Denote $\angle AFE = \alpha$ and $\angle AEF = \beta$. From this we obtain that $\alpha + \beta + 60^{\circ} = 180^{\circ}$, or $\alpha + \beta = 120^{\circ}$, implying that $\angle BFD = \alpha$ and $\angle CED = \beta$. Now from the law of sines in $\triangle DFA$ and $\triangle DEA$ we obtain that $\frac{AF}{\sin(\angle FDA)} = \frac{AD}{\sin(180^{\circ} - \alpha)}$ and $\frac{AE}{\sin(\angle EDA)} = \frac{AD}{\sin(180^{\circ} - \beta)}$, or equivalently $\sin(\angle FDA) = \frac{AF\cdot\sin(\alpha)}{AD}$ and $\sin(\angle EDA) = \frac{AE\cdot\sin(\beta)}{AD}$. Dividing these two relations we get that $\frac{\sin(\angle FDA)}{\sin(\angle EDA)} = \frac{AF}{AE}\cdot\frac{\sin(\alpha)}{\sin(\beta)} = \frac{AF}{AE}\cdot\frac{AE}{AF} = 1$. The last equality was obtained from the law of sines in $\triangle AEF$. So $\sin(\angle FDA) = \sin(\angle EDA)$ and $\angle FDA$, $\angle EDA \in [0, \frac{\pi}{3}]$, thus $\angle FDA = \angle EDA$, which implies the conclusion.