Based on your notations and what you have: $$B,A,E,F,C~{\rm are~concyclic~on~the~circumcircle~}(ABC)~{\rm with~center~}O,$$ $$AB\parallel FC~{\rm and~}AE\parallel BC.$$ Let $H$ be the orthocenter of triangle $ABC$. Let $C'$ be the alternative intersection of $CH$ with $(ABC)$. Let $A'$ be the alternative intersection of $AH$ with $(ABC)$. Then $AA'$ (resp. $CC'$) being orthogonal transversal of parallel lines $AE$ and $BC$ (resp. $AB$ and $FC$), one has $$\angle A'AE=90^\circ~({\rm resp.}~\angle C'CF=90^\circ).$$ It follows that both $A'E$ and $C'F$ are diameters of $(ABC)$, so $A'E\cap C'F=O$. Now the six points $C',A,E,A',C,F$ on circle $(ABC)$ satisfy $$CC'\cap AA'=H,C'F\cap EA'=O,AF\cap EC=D.$$ So by Pascal's theorem, $H,O,D$ are collinear. Since $HO$ coincides with the Euler line $GO$, the same is true for $G,O,D$. QED
A good diagram surely helps. First draw $\ell_1$ cutting sides $AB,BC,CA$ of $\triangle ABC$ in $X,Y,Z$ respectively. Now take points $X',Y'$ on $AB,BC$ respectively so that $XM_C=X'M_C$ and $YM_A=Y'M_A$. Then the condition that $\ell_1,\ell_2$ meet the sides symmetrically across midpoints mean, $\ell_2$ pass through $X',Y'$.
It should be considered an exercise, preliminary to current question, to show that intersection of $\ell_2$ with $AC$, $Z'$, lies symmetrically to $Z$, that is, $ZM_B=Z'M_B$.
For such a configuration, the angle between $\ell_1,\ell_2$ could vary; here it is given that $\angle(\ell_1,\ell_2)=90^\circ$. For this constraint, one has to show that their intersection point $P$ lies on the circle through the three midpoints.
It is clear that $\angle M_AM_CM_B = \angle C$. For right $\triangle ZPZ'$, $M_B$ is circumcenter and $PM_B=Z'M_B$. So $\angle M_BPZ'= \angle M_BZ'P$. Similarly $PM_AY'$ is isosceles and $\angle M_APY' = \angle M_AY'P=\angle CY'Z'$.
In $\triangle Y'CZ'$, we see the exterior angle $$\angle C = \angle CY'Z' + \angle CZ'Y' =\angle M_APY' + \angle M_BPZ' = \angle M_APM_B $$
Thus $\angle M_APM_B = \angle M_AM_CM_B$ and it follows that $P$ lies on the said circle.
Best Answer
Hint:
Prove that $\angle AFM= \angle DCM$ by considering their relationships with $\angle ABC$.