$3$ median of a triangle $ABC$ divided this triangle into $6$ part. Prove that circumcentre of each triangle is circumscribed. $3$ median of a triangle $ABC$ divided this triangle into $6$ parts. 

How do I prove that circumcentre of each triangle is circumscribed?
 A: Take the centroid ($O$) to be the origin, and place $A$ on the $x$-axis. With medians $AD$, $BE$, $CF$ (slightly different than the OP's diagram ($E$ and $F$ swapped)), we can write
$$A = -4(b+c,0) \qquad B = \phantom{-}4(b,k) \qquad C = \phantom{-}4(c,-k)$$
$$D = \phantom{-}2(b+c,0) \qquad E = -2(b,k) \qquad F = -2(c,k)$$
Consider four circumcenters associated with median $AD$. With "$XY^\perp$" abbreviating "perpendicular bisector of segment $XY$", we have that $OA^\perp$ (namely, $x=-2(b+c)$) meets $OE^\perp$ at a circumcenter $P$ with $y$-coordinate $(b^2+2bc-k^2)/k$; it meets $OF^\perp$ at circumcenter $Q$, with $y$-coordinate $-(c^2+2bc-k^2)$. The midpoint of vertical segment $PQ$ has $y$-coordinate $\frac{1}{2k}(b^2-c^2)$. This midpoint is the same as the midpoint of $RS$, where circumcenters $R$ and $S$ are the points at which $OD^\perp$ meets, respectively, $OB^\perp$ and $OC^\perp$. 
Thus, $PQ$ and $RS$ have a common perpendicular bisector, so that $PQRS$ is an isosceles trapezoid, which is necessarily cyclic. That is, the four circumcenters associated with $AD$ lie on a common circle; likewise, those associated with $BE$, as well as those associated with $CF$. This (over-)accounts for all six circumcenters.

I'm thinking there should be an elegant, coordinate-free path to the statement "$PQ$ and $RS$ have a common perpendicular bisector".
A: 
Without loss of generality, we can take the vertices of $\triangle ABC$ to be $A(6a,0),B(6b,0)$ and $C(0,6c)$ where $a<b$
So, the centroid $G(\frac{6a+6b+0}3,\frac{0+0+6c}3)$ or $(2(a+b),2c)$ 
The midpoint of $AB$ will be $D(\frac{6a+6b}2,\frac{0+0}2)$ or $(2(a+b),0)$ 
If the circum-centre of $\triangle DGB$ be $P(h,k)$
$$(h-6b)^2+(k-0)^2=\{h-3(a+b)\}^2+(k-0)^2=\{h-2(a+b)\}^2+(k-2c)^2$$
Solving the first two equation, $h=\frac{3a+9b}2$ 
Now, $h-6a=\frac{9(b-a)}2>0$ and $h-9b<0\implies 6b>h>6a$
Similarly, for $k$
