An equality from the line through the centroid of a triangle $G$ is the centroid of $\triangle\mathit{ABC}$. $X$, $Y$, and $Z$ are points on the sides (or lines through the sides) $\overline{\mathit{BC}}$, $\overline{\mathit{AC}}$, and $\overline{\mathit{AB}}$, respectively, such that $G$, $X$, $Y$, and $Z$ are collinear. If $Y$ is between $G$ and $Z$,
\begin{equation*}
\frac{1}{\mathit{GX}} = \frac{1}{\mathit{GY}} + \frac{1}{\mathit{GZ}} .
\end{equation*}
How can I show this without using vectors (directed line segments) and Menelaus' Theorem?
A solution that I have seen assumes that X is a point on $\overline{\mathit{BC}}$ and constructs points U and V that trisect the side; U is half as far from B as it is from C. (We assume that X is distinct from both U or V so that the line through G and X intersects the other two sides.) Since G is half as far from the endpoint of the median on $\overline{\mathit{BC}}$ as it is from A, and since U is half as far from B as is $\overline{\mathit{GU}}$ is parallel to side $\overline{\mathit{AB}}$. Likewise, $\overline{\mathit{GV}}$ is parallel to side $\overline{\mathit{AC}}$.
\begin{equation*}
\frac{GX}{GZ} = \frac{UX}{BU}
\qquad \text{and} \qquad
\frac{GX}{GY} = \frac{VX}{CV} .
\end{equation*}
That is fine.
Now the solution says that $\mathit{UX} = \mathit{UZ} - \mathit{XZ}$ so that
\begin{equation*}
\frac{\mathit{GX}}{\mathit{GZ}}
= \frac{\mathit{UX}}{\mathit{BU}}
= \frac{\mathit{UZ} - \mathit{XZ}}{\mathit{BU}}
\end{equation*}
and
\begin{equation*}
\frac{\mathit{UZ}}{\mathit{BU}}
= \frac{\mathit{GX}}{\mathit{GZ}} + \frac{\mathit{XZ}}{\mathit{BU}} .
\end{equation*}
Likewise, since $\mathit{VX} = \mathit{VZ} - \mathit{XZ}$,
\begin{equation*}
\frac{\mathit{VZ}}{\mathit{CV}}
= \frac{\mathit{GX}}{\mathit{GY}} + \frac{\mathit{XZ}}{\mathit{CV}} .
\end{equation*}
I would appreciate an elementary explanation of these last two equalities - involving only similar triangles and proportions - without mentioning $\mathit{UX} = \mathit{UZ} - \mathit{XZ}$ and $\mathit{VX} = \mathit{VZ} - \mathit{XZ}$.

Let $Z$ be placed on the line $AB$ such that $A$ is placed between $Z$ and $B$,  $X$ be placed on the side  $BC$, 
$Y$ be placed on the side $AC$, $BD$ be a median of $\Delta ABC$, $GX=x$, $GY=y$ and $GZ=z$.
Thus, $$[Z,Y,G,X]=[A,Y,D,C],$$ which says
$$\frac{ZG}{ZX}:\frac{YG}{YX}=\frac{AD}{AC}:\frac{YD}{YC}$$ or
$$\frac{z}{x+z}\cdot\frac{x+y}{y}=\frac{\frac{AC}{2}}{AC}\cdot\frac{YC}{YC-\frac{1}{2}AC}$$ or
$$\frac{(x+y)z}{(x+z)y}=\frac{1}{2-\frac{AC}{YC}}.$$
Now, let $GE||BC$ and $E\in DC$.
Thus, $$\frac{EC}{CY}=\frac{GX}{XY}$$ or
$$\frac{\frac{1}{3}AC}{CY}=\frac{x}{x+y},$$ which says $$\frac{AC}{CY}=\frac{3x}{x+y}.$$
Id est, 
$$\frac{(x+y)z}{(x+z)y}=\frac{1}{2-\frac{3x}{x+y}}$$ or
$$\frac{z}{(x+z)y}=\frac{1}{2y-x}$$ or
$$2yz-xz=xy+yz$$ or
$$\frac{1}{x}=\frac{1}{y}+\frac{1}{z}$$ and we are done!
Actually. By definition, for four points $A$, $B$, $C$ and $D$ which they are placed on the same line we have:
$$[A,B,C,D]=\frac{AC}{AD}:\frac{BC}{BD}.$$
You can read about this here: https://en.wikipedia.org/wiki/Cross-ratio
