Here is my over-complicated solution : Let $M_a$, $M_b$, $M_c$ and $A_1$, $B_1$, and $C_1$ be the respective midpoints and feet of altitudes in $ABC$. Let $AH$, $BH$, and $CH$ meet $(ABC)$ again at $H_a$, $H_b$, and $H_c$. Let $O_a$, $O_b$, and $O_c$ be the centers of $(CHH_a)$, $(AHH_b)$, and $(BHH_c)$. Claim: $HD_1E_1F_1$ is cyclic Notice that $E_1\in (CHH_a)$ as $\angle CE_1H=\angle ABC=\angle HH_aC$ and cyclic statements also hold. By symmetry $O_a$, $O_b$, and $O_c$ lie on $BC$, $CA$, and $AB$. To prove the claim it is sufficient to prove that the perpendicular bisectors of $HD_1$, $HE_1$, and $HF_1$ are concurrent. This is the same as having the line through $O_a$ parallel to $AC$, the line through $O_b$ parallel to $BA$, and the line through $O_c$ parallel to $CB$, are concurrent. Upon an affine transformation we see that this is equivalent to $$\frac{CO_a}{CB}+\frac{AO_b}{AC}+\frac{BO_c}{BA}=1$$However notice that $$\frac{CO_a}{CB}=\frac{\frac{1}{2}CH}{CC_1}=\frac{OM_c}{CC_1}=\frac{[AOB]}{[ABC]}$$ To finish notice that $HD_2E_2F_2$ is also cyclic and we have $$\measuredangle D_1E_1F_1=\measuredangle D_1HF_1=\measuredangle D_2HF_2=\measuredangle D_2E_2F_2$$Along with cyclic equalities, as desired.

Attachments:

...complex bash time this problem is too susceptible Set $(ABC)$ to be the unit circle and use complex numbers with $A=a$, $B=b$ and $C=c$. $D_1$ can be calculated as the chord intersection of the two lines: the line between $B$ and the antipode of $B$ and the line between $A$ and $AH\cap(ABC)$, as follows: $$d_1=\frac{\left(a-\frac{bc}a\right)(b)(-b)-(b-b)(a)\left(-\frac{bc}{a}\right)}{(b)(-b)-(a)\left(-\frac{bc}{a}\right)}=\frac{b^2(a^2-bc)}{a(b^2-bc)}=\frac{(a^2-bc)b}{(b-c)a}.$$ Similarly we can calculate $E_1$ and $F_1$ to be $$e_1=\frac{(b^2-ca)c}{(c-a)b};\quad f_1=\frac{(c^2-ab)a}{(a-b)c}.$$ Now we show $\triangle D_1E_1F_1\stackrel{-}{\sim}\triangle ABC$ by checking \begin{align*} \frac{d_1-e_1}{d_1-f_1}&=\frac{\frac{b^2(a^2-bc)(c-a)-ac(b-c)(b^2-ca)}{(b-c)(c-a)ab}}{\frac{bc(a^2-bc)(a-b)-a^2(c^2-ab)(b-c)}{(b-c)(a-b)ac}}\\ &=\frac{(a-b)(c)(a^2b^2c+a^2bc^2+ab^2c^2-a^3b^2-b^3c^2-c^3a^2)}{(a-c)(b)(a^2b^2c+a^2bc^2+ab^2c^2-a^3b^2-b^3c^2-c^3a^2)}\\ &=\frac{ac-bc}{ab-bc}\\ &=\frac{\overline a-\overline b}{\overline a-\overline c}\\ &=\overline{\frac{a-b}{a-c}}. \end{align*} Hence we conclude similarly that $\triangle D_2E_2F_2\stackrel-\sim\triangle ABC$, and therefore $\triangle D_1E_1F_1\stackrel+\sim\triangle D_2E_2F_2$.

As in #2 it is enough to prove that $H \in (D_1E_1F_1)$. I will prove it from "Ptolemy-Sinus Lemma". By Sinus Theorem in triangle $AHF_1$, $HF_1 \cdot \sin D_1HE_1 = AH \cdot \sin (B-C) = 2 R \cos A \sin (B-C) = -2R \cos (B+C) \sin (B-C)$. By symmetry it is enough to prove that $$ \sum_{cyc} \cos (B+C) \sin (B-C) = 0$$This is obvious by $ \sin X \cos Y = \frac{1}{2} \cdot (\sin (X-Y) + \sin (X+Y))$.