solution using synthetic geometry I managed to solve this problem only using complex numbers but I'd like to solve it using synthetic geometry and I can't.
Can someone help me to solve this problem using  synthetic geometry?

Let $ABC$  an acute triangle with $AB &gt; AC$ . Let $O$ its circumcenter and let $D$ the midpoint of $BC$. The circle of diameter $AD$ intersects again  $AB$ and $AC$ in $E$ and in $F$, respectively. Let $M$ the midpoint of $EF$.
Prove that $MD$ is parallel to $AO$.

This is my solution. But, as I wrote above, I'd like to solve it using synthetic geometry and I can't.

Setting the origin of the plane in O and the points A, B and C on the circumference of unit radius, we have  that  

$ a \bar{a} = 1 \text{;} \ b \bar{b} = 1 \text{;} \ c \bar{c} = 1 $.

Because $ D $ is the midpoint of $ BC $, we can write 

1) $ d = \dfrac{b + c}{2} $

and because $ AD $ is a diamtere of the new circle then said $ Q $ his midpoint we have

2) $ q = \dfrac{a + d}{2} = \dfrac{2a + b + c}{4}$ 

Said $ M_{1} $ the projection of $ Q $ on $ AB $, then 

$ m_{1} = \frac{1}{2} \left[ \left( \dfrac{\bar{q} - \bar{a}}{\bar{b} - \bar{a}}\right) (b - a) + a + q \right] $ 

but

$\dfrac{1}{\bar{b} - \bar{a}} = \dfrac{1}{\dfrac{1}{b} - \dfrac{1}{a}} = \dfrac{ab}{a – b} $

and so 

$ m_{1} = \frac{1}{2} \left[ \bar{q} ab (-1) - \dfrac{ab}{a} (- 1)  + a + q \right] \Rightarrow $

$m_{1} = \frac{1}{2} \left( a + b + q - ab\bar{q} \right) $

In the same way, said $ M_{2} $ the projection of $ Q $ on $ AC $ we have

$ m_{2} = \frac{1}{2} \left[  \dfrac{\bar{q} - \bar{a}}{\bar{c} - \bar{a}} (c - a) + a + q \right] \Rightarrow $

$ m_{2} = \frac{1}{2} \left( a + c + q - ac\bar{q} \right) $

Because $ Q $ is the center of the new circle passing through $ A \text{,} D \text{,} E \text{,} F $ we have that $ M_{1}Q $ is axes of $ AE $ and that $ M_{2}Q $ in axes of $ AF $.

So we have that $ M_{1} $ is midpoint of $ AE $ and that $ M_{2} $ is midpoint of $ AF $.
So we can write

$ m_{1} = \dfrac{a + e}{2} \ \Rightarrow \ e = 2m_{1} – a $

and also

$ m_{2} = \dfrac{a + f}{2} \ \Rightarrow \ f = 2m_{2} - a$

The point $ M $ is defined as midpoint of $ EF $  so we have 

$ m = \dfrac{e + f}{2} = \dfrac{2m_{1} + 2m_{2} - 2a}{2} = m_{1} + m_{2} - a$

therefore replacing $ m_{1} $ and  $ m_{2} $ we have

$m = \frac{1}{2} a + \frac{1}{2} c + \frac{1}{2} q + \frac{1}{2} a + \frac{1}{2} b + \frac{1}{2} q - \frac{ab\bar{q}}{2} - \frac{ac\bar{q}}{2} - a \ \Rightarrow $

$m = \dfrac{b + c}{2} + q - a\bar{q} \dfrac{b + c}{2}$

We know, furthemore, that $ \bar{q} $ is:

3) $\bar{q} = \dfrac{2\bar{a} + \bar{b} + \bar{c}}{4} = \frac{1}{4} \left( \frac{2}{a} + \frac{1}{b} + \frac{1}{c} \right) = \dfrac{2bc + ab + ac}{4abc}$

Now we write the equation of the parallel line through $D$ parallel to $AO$:

let $z$ be a generic point of this line it is possible to write

4) $ \dfrac{z - d}{a - 0} = \dfrac{\bar{z} - \bar{d}}{\bar{a} – 0} $

Replacing $d$ with the expression of 1) and taking into account that $ \bar {a} = \frac{1}{a} $, we get

5) $z - \dfrac{b + c}{2} = a^{2} \left( \bar{z} - \dfrac{\bar{b} + \bar{c}}{2} \right) = a^{2}\bar{z} - \dfrac{a^{2} (\bar{b} + \bar{c})}{2}$

If the $ M $ point belongs to this line, $ m $ must satisfy equation 5). Substituting the value of $ m $ we have

$\dfrac{b + c}{2} + q - a \bar{q} \cdot \left( \dfrac{b + c}{2} \right)  - \dfrac{b + c}{2} = a^{2}\left( \dfrac{\bar{b} + \bar{c}}{2} + \bar{q} - \bar{a}q \cdot \dfrac{\bar{b} + \bar{c}}{2} \right) - \dfrac{a^{2} (\bar{b} + \bar{c})}{2}   \Rightarrow $

$ q - a \bar{q} \cdot \left( \dfrac{b + c}{2} \right) = a^{2}\bar{q} - \dfrac{aq( b + c)}{2bc} \ \Rightarrow $

$ q \left( 1 + \dfrac{ab + ac}{2bc} \right) = \bar{q}a \left( a + \dfrac{b + c}{2} \right)$ 

Substituting the values of  $ q $ in 2) and the value of $ \bar{q} $ in 3) we have 

$ \dfrac{(2a + b + c)(2bc + ab + ac)}{8bc} = \dfrac{a (2bc + ab + ac) (2a + b + c)}{a \cdot 8bc} $

which is, obviously, an identity, so $ MD $ is parallel to $ AO $, as we wanted to prove.
Hints:

Construct squares $EDPR$ and $FDQS$ outwardly on the sides $ED$ and $FD$ of $\triangle DEF$. Let the line $PQ$ intersect lines $AB, AC$ at $U, V$ respectively .

First show that $MD \perp \operatorname{line} PQ = \operatorname{line}\ UV$. Therefore if $l$ is the line through $A$ that is parallel to $MD$ then $l \perp \operatorname{line} UV$.

Using $\triangle DEF \cong \triangle ABC$ (use ratio $AB/AC$ and $\angle A$) and $\operatorname{line} DP \parallel \operatorname{line} AB$ show that $\angle AUV = \angle ACB$. Since $\angle OAB = \frac\pi 2 - \angle ACB$, conclude that $\operatorname{line} AO \perp UV$.

Thus lines $AO$ and $l$ through $A$ are both perpendicular to the line $UV$, which implies that $\operatorname{line} AO = l \parallel \operatorname{line} MD$.
I was not able to solve it using synthetic geometry either. Is there anyone who can solve it and post a solution using synthetic geometry methods?