Show that 3 points are collinear I'm struggling to solve this problem geometrically,I solved it with coordinates and now Im trying to prove it geometrically,but i came to no conclusion.
In an equilateral triangle we have the orthocentrum(O) and we choose a random point(M),then from that point we draw perpendiculars to each side of the triangle.The points that these perpendiculars meet with the triangle will make a new triangle lets say EFD.In this new EFD triangle we will have a new point (G) that is the centroid of triangle EFD.I need to show that M,G and O are collinear.
Here is the figure to this problem that I did in Geogebra.

 A: Here is another solution which can be considered geometric on one hand but as well analytical on the other hand.
Knowing Viviani's theorem, we can assume WLOG that equilateral triangle $ABC$ has sides such that the sum of distances to the (constant) sum of the sides is equal to $1$.
I assume you know barycentric coordinates (or trilinear coordinates which are the same for such a reference triangle).
I will take notation "b.c." for "barycentric coordinates".
I will use notations $A'B'C'$ for the so-called pedal triangle associated with $M$ (instead of $DEF$ in order to have a better homogeneity) ; see below:

On this figure, if the b.c. of $M$ are $(a,b,c)$ with normalization:
$$a+b+c=1\tag{1}$$
then, in particular, the b.c. of the foot $A'$ of the perpendicular issued from $M$ onto $BC$ are:
$$A'(0,\ \ \underbrace{b+\tfrac12 a}_{\text{see fig.}}, \ \ c+\tfrac12 a)\tag{2}$$
with a graphical proof for the coordinate $b+\tfrac12 a$ (factor $\tfrac12$ coming from the fact that $\cos \angle HA'M=\cos \pi/3=\tfrac12$).
For similar reasons, the b.c. of $B'$ and $C'$ are:
$$B'(a+\tfrac12 b, \ \ 0, \ \ c+\tfrac12 b) \ \text{and} \ C'(a+\tfrac12 c, \ \ b+\tfrac12 c, \ \ 0)\tag{3}$$
Therefore, the b.c. of the centroid $G$ of $A'B'C'$ are, using (2) and (3) :
$$G(\tfrac13(2a+\tfrac{b+c}{2}), \ \ \tfrac13(2b+\tfrac{a+c}{2}), \ \ \tfrac13(2c+\tfrac{a+b}{2}))\tag{4}$$
Using (1), we can transform the b.c. in (4) into
$$G \left(\frac{3a+1}{6}, \ \ \frac{3b+1}{6}, \ \ \frac{3c+1}{6}\right)$$


*The determinant of b.c. of $M,O,G$
$$\begin{vmatrix}a&b&c\\
1/3&1/3&1/3\\
(3a+1)/6&(3b+1)/6&(3c+1)/6
\end{vmatrix}=\frac{1}{18}\begin{vmatrix}a&b&c\\
1&1&1\\
(3a+1)&(3b+1)&(3c+1)
\end{vmatrix}$$
is clearly zero, proving the alignment of points  $M,O,G$.

Moreover the b.c. of $G$ being the half-sum of the b.c. of $M$ and $O$, we can conclude that $G$ is the midpoint of $M$ and $O$.
