Given $A(a,b),B(0,0),C(c,0)$. Then $D(\frac{2a+c}{3},\frac{2b}{3})$ and $E(\frac{2a+c}{3},0)$. Point $F((\frac{2a+c}{6},\frac{(2a+c)\sqrt{3}}{6})$, point $F \in BD$ and point $F \in AE$, giving $c=-8a,b=-3\sqrt{3}a$.

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Other solution?

VicKmath7 wrote: Other solution? Let $\angle EAD=\alpha$. By Law of sines in triangle $ADE$ we have $\frac{DE}{sin\alpha}$$=$$\frac{AD}{sin30}$=$\frac{AD}{1/2}$$=$$2AD$$=$$DC$, which implies that $sin\alpha$=$\frac{DE}{DC}$=$sin\angle ACE$. Now $\alpha + \angle ACE$ cannot equal to $180$ because they are angles in triangle $AEC$, impiles that $\alpha$=$\angle ACE$ implies $\alpha + \angle ACE = \angle AEB = 60$. So $2\alpha=60$ gives $\alpha = 30$. In triangle $ADF$ we have $\angle ADF = 180 - \alpha - \angle AFD=180 - 30 - 60= 90$.