Geometry- Circle Theorem In the given figure, $BH\perp AD$, $DC\perp AB$, $AF\perp BD$ and $DE=EB$ then prove that $AG$ is the diameter of the circle.

My Attempt: 
$\angle AHB=90$ and $\angle ACD=90$.
Let $BH$ and $CD$ intersects at $P$. Then, $\ ACPH$ is a cyclic quadrilateral. 
Please, Help me to complete from here.
 A: 
Let's call $P$ the orthocenter. Now we are going to prove that the symmetric point of $P$ w.r.t $E$ (let's call him $N$) lies on the circumcircle.
The quadrilateral $PBND$ is a parallelogram because both diagonals cut themself in the midpoint. Then $\angle DPB =\angle DNB$ but we know that $\angle DPB=\angle A -180º$ and so $\angle DNB=\angle A-180º$. That give us the quadrilateral $ABND$ is cyclic and this circle is, of course, the circumcircle. 
Once $N$ and $G$ lies on the same line and on the same circle then $N=G$ and then $PBGD$ is a parallelogram. Now we have that $BH \parallel DG$ and once $BH \perp AD$ then $DG \perp AD$ what give us that $AG$ is the diameter of the circle.  
P.S: Why $\angle DPB=\angle A -180º$?
In the triangle $AHB$ we have $\angle ABH = 90º-\angle A$ (becasue $\angle AHB =90º$).
Then at the triangle $CPB$ we have $\angle CPB =\angle A$ (because $\angle PCB=90$ and $\angle CBP = \angle ABH = 90º-\angle A$) and $\angle CPB + \angle DPB = 180º$.
A: I have something close:
The point where the altitudes intersect is the orthocenter, lets call it $O$
I would like to define $G$ as the point of intersection of $OE$ and $AM$ and then show that that $G$ lies on the circle.
The point where the medians intersect is the centroid -- $N$
The center of the cirlce is the circumcenter -- $M$

$ME \parallel AF$  Not sure if I actually need this.  But $M$ is at the intersection of the perpendicular bisectors of $ABC$  
$O,N,M$ lie on a line called the Euler line
$ON:NM = 2:1$
$AN:NE = 2:1$
$N$ is also the centroid of $AGO$
$AO$ and $AE$ are medians of $AGO$
$ME$ is the midline of $AGO$
$AM = MG$
$G$ lies on the circle centered at $M.$
