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 $AB > AC$ . 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.
 A: 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 $\operatorname{line} AO = l \parallel \operatorname{line} MD$.
A: 
Considering two cases where only one of the two conditions is satisfied,
first let $ABC$ be an acute triangle but with $AB=AC$, as in the figure above. Since $MD$ and $AO$ are collinear, they are parallel in that they do not intersect.

Next, suppose $AB>AC$, but that the angle at $C$ is right. Then since $AD$ is the diameter of the lesser circle, $\angle AFD$ is right. But $\angle ACB$ is also right. Therefore $F$ and $C$ coincide, and in the triangle on base $EB$, which lies on $AO$ extended,$$\frac{FM}{ME}=\frac{CD}{DB}$$ making $MD$ again parallel to $AO$.

Alternatively, keeping $O$ within $\triangle ABC$ so that $\angle ACB$
is acute, and with $J$ as the midpoint of $AD$, if we move $C$ toward $A$ until $\angle CAB$ is right, then $CB$ is a diameter, so that $D$ coincides with $O$, and $M$ with $J$ and $H$. And since triangles $AME$ and $ADB$ are isosceles, then$$\angle DME=2\angle MAE$$and$$\angle CDM=2\angle DBA$$But $\angle DBA=\angle MAE$. 
Therefore $$\angle CDM=\angle DME$$and $FE$ is parallel to $CB$.  
Note that in all of these cases the angle between $CB$ and $FE$ equals $\angle OAD$ between the diameters. This is clearly $0^o$ in the first case, and in the second case (second figure) $\angle EAD$ and $\angle ECD$ at the circumference stand on common arc $ED$ and hence are equal. The alternative second case (third figure) is like the first case. 
In cases where both conditions are satisfied, that is where $$\angle ABC<\angle ACB<90^o$$$EF$, $BC$ intersect at some point $G$, with $\angle BGE$ again equal to the angle  of the diameters.
In the fourth figure, triangles $GDH$ and $DMH$ have the angle at $H$ in common. 
And $$\angle GDH=\angle DMH$$ This was true in the first case above, where $\angle CDM=\angle DME$. 
And in the second case, second figure, since$$\angle DCE=\angle DAE=\angle MDA$$then triangles $CDH$ and $DMH$ are similar, making $\angle CDH=\angle DMH$. 
In the alternative second case (third figure) again $\angle CDM=\angle DME$. 
Since in the general case (fourth figure) $\angle GDH=\angle DMH$, and the angle at $H$ is shared, then $$\angle DGH=\angle MDH$$ 
But $\angle DGH$ is equal to the angle of the diameters.  
Therefore $$\angle MDH=\angle OAD$$ and $MD$ is parallel to $AO$.
[The foregoing is the nearest I get to formal synthetic proof so far. 
 Not sure about the validity here of using borderline cases to argue the  general case. Suggestions welcome.]         
A: 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?
