Based on the described construction and using the OP's notations, we have the following figure:
It is enough to demonstrate that $DG=r$, the radius of the circle.
By the construction, the triangle $ABD$ is equilateral and then $AB=BD=AF$. $BDG$ is an isosceles triangle because the triangles $ABG$ and $ADG$ are congruent. $\angle AGD = \angle BGA$ and $AB=AD=AF$.
Now, connecting $A$ and $F$ with the center $O$ we get the green triangle $AFO$. To show that $DG=FO=r$, it is enough to prove that the triangles $BDG$ and $AFO$ are congruent.
We have already seen that the bases of these triangles equal. What remains is to show that $\angle AOF = \angle BGD$.
This is easy. Just look at the figure and you will see that the $\angle BDG =2\times\angle AGD = \angle AOF$. (See the Wiki article called Inscribed angle.)
Finally, this is how to use the construction at stake to find the center of the circle:
- Construct $E$, $F$, and $D$ and then $G$ as described.
- The straight $ED$ and the circle centered at $G$ through $D$ will meet in $O$.
Note
This construction is not that bad compared to the most frequently used one in the case of which you construct two bisecting perpendiculars. In the case of this present construction it is enough to draw the line through $F$ and $D$ which already exist after constructing the first bisecting perpendicular.
The nine point circle exists in all triangles , but for this answer , we will be focusing on the case when $\triangle ABC$ is acute .
First , let us remember the properties of cyclic quadrilaterals , as this will help us prove the concyclicity of points .
The main points to remember , are that the opposite angles are supplementary , and that the chords subtend equal angles at the circumference . The equal point to note is that the converse is true as well. This is how we will be proving that the points lie on a circle.
For the problem , it may be best to solve it in steps .
Step $1$:-
In step $1$ , we can perhaps prove that the feet of the altitudes lie on the same circle as the midpoints. 
To do this , a single altitude should suffice . As, if we can prove that this lies on the same circle as the midpoints , the others must as well. To prove this , maybe the midpoint theorem , and basic angle chasing will help...
Hint:-
Prove $\triangle FAX$ is isosceles using the converse of the midpoint theorem . Then prove that $\angle FXE$ and $\angle FDE$ are supplementary .
Step $2$:-
I assume you have finished step $1$ . Now , we use the information that the feet of the altitudes and the midpoints lie on the same circle , to prove that the midpoints of the lines joining the vertices to the orthocentre also lie on the same circle .
Again , properties of altitudes and midpoints should help us solve this . Note that in the figure , $M’$ is the midpoint of the line joining vertex $C$ to orthocentre $H$ . $M$ is the midpoint of $BC$
Hint:-
$M’M$ joins the midpoints of two line segments ! Use the midpoint theorem . Also , make a great observation. Since $\angle CFB = \angle BEC = 90 $ , $CB$ is the diameter of the circle which inscribes cyclic quadrilateral $CEFB$ ! Also , $M$ is the midpoint of the diameter... By angle chasing , and using these observations , you should be able to prove that $\angle FM’M = \angle FEM$, proving that the quadrilateral is cyclic
This completes our proof , as it must follow that all the other midpoints of the lines joining the vertices with the orthocentre are concyclic as well.
Ofcourse , we have proved the existence of the $9$-point circle for only acute angled triangles , but I trust you should now be able to prove it for all other triangles as well!
Best Answer
It is possible to define proportionality relationships between line segments in purely geometric terms. The trick is that you force the line segments to be legs of a right triangle, which makes everything well-defined.
More formally, we say that $AB:AC=AD:AE$ if in the following diagram the hypotenuse line segments $BC$ and $DE$ are parallel:
If $p,q,r,s$ are arbitrary line segments, then we can build a diagram like the above one using segments congruent to $p,q,r,s$ as legs of the two right triangles. We then say that $p:q=r:s$ if that diagram has parallel hypotenuses.
You can then define multiplication purely geometrically as well: you choose some particular line segment $1$, and for any segments $p$ and $q$ you say that the line segment $pq$ has length such that $1:q=p:pq$.
Once you have these definition, you can prove that in fact the definition $p:q=r:s$ has all the standard proportionality properties (e.g., that two triangles are similar if and only if their corresponding sides are proportional), and that the above multiplication forms a field along with the obvious addition operation on segment lengths by concatenation. These proofs are complicated and a little tedious but basically boil down to applying the inscribed angle theorem for circles lots of times in clever ways. You can find details in Hilbert's original book (though the Gutenberg version is kind of error-prone, so check his proofs carefully) or in Hartshorne's Euclidean geometry book.