There are a number of results in triangle geometry of the type: if two specific centres (as a concrete example, the incentre and the circumcentre) coincide, then the triangle is equilateral. Does anybody know of a synthetic proof for the corresponding result for the centroid and the Feuerbach centre? (for definitions, I refer to the easily accessed site "Encyclopedia of Triangle Centers").
[Math] Coincidence of triangle centres
geometry
Related Solutions
Here's a (not-to-scale) picture of the situation:
Necessarily, each circle center ($D$, $E$, or $F$) is the point where an angle bisector meets an opposite edge; moreover, the points of tangency of a circle with the adjacent edges (for instance, $D^\prime$ and $D^{\prime\prime}$) are simply the feet of perpendiculars from the center to those edges.
We'll write $a$, $b$, $c$ for the lengths of sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$, and $d$, $e$, $f$ for the radii of $\bigcirc D$, $\bigcirc E$, $\bigcirc F$. Now, observing that each angle bisector cuts the triangle with into sub-triangles with convenient "bases" and "heights", we can compute the area, $T$, of $\triangle ABC$ in three ways:
$$T \;=\; \frac12 d\;(b+c) \;=\; \frac12 e\;(c+a) \;=\; \frac12 f\;(a+b) \tag{1}$$
Of course, writing $r$ for the inradius of $\triangle ABC$, we have a well-known fourth formula for area: $$T \;=\; \frac12 r\;(a+b+c) \tag{2}$$
We can easily eliminate $a$, $b$, $c$ from the above. For instance, $$\begin{align} b+c = \frac{2T}{d}\quad c+a=\frac{2T}{e}\quad a+b = \frac{2T}{f} &\quad\to\quad 2(a+b+c) = 2T\left(\;\frac{1}{d}+\frac{1}{e}+\frac{1}{f}\;\right) \tag{3} \\ a+b+c = \frac{2T}{r} &\quad\to\quad 2(a+b+c) = 2T\left(\;\frac{2}{r}\;\right) \tag{4} \end{align}$$ so that, as @Jack notes,
$$\frac{2}{r} = \frac{1}{d} + \frac{1}{e} + \frac{1}{f} \tag{$\star$}$$
Addendum (four years later!). As @jmerry has observed, the specific configuration in the problem statement is invalid. If we solve $(1)$ for $a$, $b$, $c$, we find $$a = \left(-\frac1d + \frac1e + \frac1f \right)T \qquad b = \left(\frac1d - \frac1e + \frac1f \right)T \qquad c = \left(\frac1d + \frac1e - \frac1f \right)T$$ With $d=18$, $e=6$, $f=9$, these become $a=2T/9$, $b=0$, $c=T/9$, which do not correspond to a valid triangle ... not even a validly-degenerate one. (It's a good thing I didn't claim my picture was to scale.) A valid triangle requires that the three aspects of the Triangle Inequality hold $$a \leq b+c \qquad b \leq c+a \qquad c \leq a+b$$ which, in turn, require $$\frac3d \geq \frac1e + \frac1f \qquad \frac3e \geq \frac1f+\frac1d \qquad \frac3f \geq \frac1d+\frac1e$$ (The first of these is violated by the given values of $d$, $e$, $f$.)
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!
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Best Answer
The Feuerbach center is the midpoint of the orthocenter and the circumcenter (see e.g. Theorem 1.82 of Coxeter-Greitzer, Geometry Revisited, page 21).
The centroid divides the segment from the orthocenter to the circumcenter in the ratio $2 : 1$.
Therefore the Feuerbach center, the centroid, the orthocenter and the circumcenter must coincide.
A triangle in which any two of circumcenter, orthocenter and centroid coincide is equilateral.