<s>[INST] How find this $DF=?$ In diamond $ABCD$,such $\angle B=\dfrac{\pi}{3}$,and the point $E$ in on $BC$.such
$BE=3CE$,and the point $F$ is on $DE$,such $\angle AFC=\dfrac{2\pi}{3}$
Find $$DF=?$$

My try: since
$$\angle B+\angle AFC=\pi$$
so
$A,B,C,F$  is cyclic
and follow I can't
 [/INST] Let's draw another equilateral triangle $BCG.$
Let $O$ be the center of that triangle.
Let's show that points $D, E$ and $O$ are collinear $\Longleftrightarrow$ points $D, F$ and $O$ are collinear.
$\overrightarrow{DE}=\overrightarrow{DC}+\overrightarrow{CE}=\overrightarrow{DC}+\dfrac{1}{4}\overrightarrow{CB},$
$\overrightarrow{DO}=\dfrac{1}{3}\left(\overrightarrow{DB}+\overrightarrow{DC}+\overrightarrow{DG}\right) = \dfrac{1}{3}\left(\overrightarrow{DA}+\overrightarrow{AB}+\overrightarrow{DC}+2\overrightarrow{DC}\right)=\dfrac{1}{3}\left(\overrightarrow{CB}+4\overrightarrow{DC}\right) = \dfrac{4}{3}\overrightarrow{DE}.$
Thus, $D, F, E$ and $O$ are collinear.

Since $\angle COB + \angle BAC = \dfrac{2\pi}{3} + \dfrac{\pi}{3} = \pi,$ all five points $B, A, F, C$ and $O$ are concyclic.
The circumscribed circle of $\triangle ABC$ is tangent to line $DC$, which follows from the fact that $\angle BAC = \dfrac{\pi}{3} = \angle BCG.$ As an alternative, it can be observed from the symmetry of the picture with respect to the perpendicular bisector of $AB.$
It follows from the tangent-secant theorem that $DC^2 = DF\cdot DO.$ We know $DC,$ we want $DF,$ so the only thing that's left to calculate is $DO.$
We use cosine formula for that. Assuming the side of any of the equilateral triangles is $a$, $DO^2 = CD^2+CO^2-2\cdot CD \cdot CO \cos \cfrac{5\pi}{6}=a^2+\dfrac{a^2}{3} + 2a\dfrac{a}{\sqrt{3}}\dfrac{\sqrt{3}}{2} = \dfrac{7}{3}a^2.$
So $DF = a^2\bigg/\sqrt{\dfrac{7}{3}a^2}=\sqrt{\dfrac{3}{7}}a = \sqrt{\dfrac{3}{7}}AB$
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