Triangles and medians Let G be the ABC barycenter. A line intersects the medians AD, BE and CF in X, Y and Z, respectively. Prove that
$$\frac{XD}{XG}+\frac{YE}{YG}+\frac{ZF}{ZG}=3$$
By areas relations I found that
$\frac{XG\cdot YG}{BG\cdot DG}=\frac{6SG_{\triangle XYG}}{S_{\triangle ABC}}\\
\frac{XG\cdot ZG}{CG\cdot DG}=\frac{6SG_{\triangle XZG}}{S_{\triangle ABC}}\\
\frac{YG\cdot ZG}{BG\cdot CG}=\frac{3SG_{\triangle YZG}}{S_{\triangle ABC}}$
And using that
$S_{GYZ}=S_{GXZ}+S_{GXZ}$
I wrote
$$XG\cdot YG\cdot CG+XG\cdot ZG\cdot BG=2YG\cdot ZG\cdot DG$$
But I can't solve this problem.
Can someone help me?
Thanks for antetion!

Without loss of generality, we can scale and place the triangle so that its center of mass is in the origin and one of its vertices is on the $y$-axis at a unitary distance from the center of mass. So the vertices are $A(0,1)$, B$(p,q)$, and $C(-p,r)$. By well known properties of the medians, the midpoints of $BC$, $AC$, and $AB$ are, respectively:
$$D\Big(0, -\frac{1}{2}\Big)$$
$$E\Big(-\frac{p}{2}, -\frac{q}{2}\Big)$$
$$F\Big(\frac{p}{2}, -\frac{r}{2}\Big)$$

Also note that, because the $y$-coordinate of $D$ is the average  of the $y$-coordinates of $B$ and $C$, we have $(q+r)/2=-1/2$ and then $q+r=-1$.
The medians correspond to the lines
$$AD\rightarrow\,\,\, x=0$$
$$BE\rightarrow\,\,\, y=\frac{q}{p}\,x$$
$$CF\rightarrow\,\,\, y=-\frac{r}{p}\,x  $$
Now let us draw a line $y=mx+n$, not parallel to any of the medians.  Its intersections with the medians $AD$, $BE$, and $CF$ are, respectively
$$X\left(0,n \right)$$
$$Y\left(\frac{np}{q-mp},  \frac{nq}{q-mp} \right)$$
$$Z\left(-\frac{np}{r+mp},  \frac{nr}{r+mp}\right)$$
Then we have
$$\frac{XD}{XG}=\left|\frac{n+1/2}{n}\right|=1+\frac{1}{2n}$$
$$\frac{YE}{YG}=\frac{ \sqrt{ 
      \left(     \frac{np}{q-mp}+\frac{p}{2}
      \right)^2+\left(    
  \frac{nq}{q-mp}+\frac{q}{2}     \right)^2     }}{   \sqrt{ 
      \left(     \frac{np}{q-mp}
      \right)^2+\left(    
  \frac{nq}{q-mp}\right)^2      }     }\\
=\sqrt{1+\frac{(np^2+nq^2)(q-mp)  }{n^2(p^2+q^2)}+\frac{(p^2+q^2)(q-mp)^2}{4n^2(p^2+q^2)}
   }\\
\sqrt{ 1+\frac{q-mp}{n}+ \frac{(q-mp)^2}{4n^2} }\\
= 1+\frac{q-mp}{2n} $$
$$\frac{ZF}{ZG}=\frac{ \sqrt{ 
      \left(     \frac{-np}{r+mp}-\frac{p}{2}
      \right)^2+\left(    
  \frac{nr}{r+mp}+\frac{r}{2}     \right)^2     }}{   \sqrt{ 
      \left(     \frac{-np}{r+mp}
      \right)^2+\left(    
  \frac{nr}{r+mp}\right)^2      }     }
\\
=\sqrt{1+\frac{(np^2+nr^2)(r+mp)  }{n^2(p^2+r^2)}+\frac{(p^2+r^2)(r+mp)^2}{4n^2(p^2+r^2)}
   }\\
\sqrt{ 1+\frac{r+mp}{n}+ \frac{(r+mp)^2}{4n^2} }\\
= 1+\frac{r+mp}{2n} 
$$
Therefore
$$\frac{XD}{XG}+\frac{YE}{YG}+\frac{ZF}{ZG}\\= 1+\frac{1}{2n}+1+\frac{q-mp}{2n}  + 1+\frac{r+mp}{2n} \\
= 3+\frac{1}{2n}+\frac{q+r}{2n}  $$
and since $q+r=-1$, we get
$$\frac{XD}{XG}+\frac{YE}{YG}+\frac{ZF}{ZG}= 3$$
