This problem has several different proofs. The trigonometry based approach requires some computation, but is one of the most straightforward solutions. For convenience, let $\angle{A} = 3\alpha, \angle{B} = 3\beta,$ and $\angle{C} = 3\gamma$. Furthermore, let $R$ be $\triangle{ABC}$'s circumradius. Then apply the law of sine: \begin{align*} BD &= \frac{BC}{\sin\angle{BDC}}\cdot\sin\angle{BCD}\\ &=\frac{2R\sin 3\alpha}{\sin(180^\circ - \beta - \gamma)}\cdot\sin\gamma\\ &=2R\cdot\frac{\sin 3\alpha \cdot\sin\gamma}{\sin(\beta + \gamma)}\\ &=2R\cdot\frac{4\sin\alpha\sin(60^\circ+\alpha)\sin(60^\circ-\alpha)\sin\gamma}{\sin{(\beta+\gamma)}}&\because\scriptsize{\text{\myRefEquation{eq_trig_triple_s_2}}}\\ &=8R\cdot\sin\alpha\sin(60^\circ+\alpha)\sin\gamma \end{align*} By symmetry, it must hold that $$BF=8R\cdot\sin\gamma\sin(60^\circ+\gamma)\sin\alpha$$ Therefore by the law of cosine, we have \begin{align*} &DF^2 \\ &= BD^2 + BF^2 - 2\cdot BD\cdot BF\cos\angle{DBF}\\ &= 64R^2\sin^2\alpha\sin^2(60^\circ+\alpha)\sin^2\gamma\\ &\quad+64R^2\sin^2\gamma\sin^2(60^\circ+\gamma)\sin^2\alpha\\ &\quad-2\cdot 8R\sin\alpha\sin(60^\circ+\alpha)\sin\gamma \cdot 8R\sin\gamma\sin(60^\circ+\gamma)\sin\alpha\cdot\cos\beta\\ &=64R^2\cdot\sin^2\alpha\sin^2\gamma(\sin^2(60^\circ+\alpha)+2\sin^2(60^\circ+\gamma)\\ &\quad-\sin(60^\circ+\alpha)\sin(60^\circ+\gamma)\cos\beta) \end{align*} It is possible to simplify the expression inside the bracket using trigonometric identities. However, there is a simpler way to show that it equals $\sin^2\beta$. Given $(60^\circ+\alpha) + (60^\circ+\gamma) + \beta = 180^\circ$ and $\alpha, \beta, \gamma > 0$, there must exist a triangle with angles $(60^\circ+\alpha), (60^\circ+\gamma),$ and $\beta$. Let its sides be $x, y,$ and $z$, respectively, and $R'$ be its circumradius. Then apply the law of cosine and the law of sine: \begin{align} z^2 &= x^2 + y^2 - 2xy\cos\beta\\ (2R'\sin\beta)^2 &= (2R'\sin(60^\circ+\alpha))^2+(2R'\sin(60^\circ+\gamma))^2\\ &\quad - 2(2R'\sin(60^\circ+\alpha))(2R'\sin(60^\circ+\gamma))\cos\beta\\ \sin^2\beta &= \sin^2(60^\circ+\alpha)+2\sin^2(60^\circ+\gamma)\\ &\quad-\sin(60^\circ+\alpha)\sin(60^\circ+\gamma)\cos\beta \end{align} Setting this result to the previous equation leads to $$DF^2=64R^2\sin^2\alpha\sin^2\beta\sin^2\gamma\implies DF=8R\sin\alpha\sin\beta\sin\gamma$$ Similarly, we must have $$DE=EF=FD=8R\sin\alpha\sin\beta\sin\gamma$$