This follows from the Intercept Theorem .
Let $DE$ be the line segment joining the midpoint $D$ of $AB$ and the midpoint $E$ of $BC$.
Draw a line parallel to to $DE$ that passes through $A$. Extend the side $BC$ so that it intersects this line in the point $F$. By the intercept theorem,
$$
{DB\over DA}={BE\over EF}
$$
But $DB=DA$, so, $EF=BE=EC$. It follows that $F=C$, and, thus, $AC$ is parallel to $DE$.
(Of course, you could argue using similar triangles too. The intercept Theorem is equivalent to "the similar triangle business".)
a single line through the centroid can only intersect all three sides of a triangle at $X,Y,Z$ if two of these points are coincident at a vertex of the triangle, and the third is the midpoint of the opposite side
The sides of the triangle are lines of infinite length in this context, not just the line segments terminated by the corners of the triangle. OK, Andreas already pointed that out in a comment, but since originally this was the core of your question, I'll leave this here for reference.
I recently posted an answer to this question in another post. Since @Michael Greinecker♦ asked me to duplicate the answer here, that's what I'm doing.
I'd work on this in barycentric homogenous coordinates. This means that your corners correspond to the unit vectors of $\mathbb R^3$, that scalar multiples of a vector describe the same point. You can check for incidence between points and lines using the scalar product, and you can connect points and intersect lines using the cross product. This solution is influenced heavily by my background in projective geometry. In this world you have
$$
A = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \qquad
B = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \qquad
C = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \\
G = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \qquad
l = \begin{pmatrix} 1 \\ a \\ -1-a \end{pmatrix} \qquad
\left<G,l\right> = 0 \\
X = \begin{pmatrix} a \\ -1 \\ 0 \end{pmatrix} \qquad
Y = \begin{pmatrix} 1+a \\ 0 \\ 1 \end{pmatrix} \qquad
Z = \begin{pmatrix} 0 \\ 1+a \\ a \end{pmatrix}
$$
The coordinates of $l$ were chosen such that the line $l$ already passes through $G$, as seen by the scalar product. The single parameter $a$ corresponds roughly to the slope of the line. The special case where the line passes through $A$ isn't handled, since in that case the first coordinate of $l$ would have to be $0$ (or $a$ would have to be $\infty$). But simply renaming the corners of your triangle would cover that case as well.
In order to obtain lengths, I'd fix a projective scale on this line $l$. For this you need the point at infinity on $F$. You can obtain it by intersecting $l$ with the line at infinity.
$$ F = l\times\begin{pmatrix}1\\1\\1\end{pmatrix}
= \begin{pmatrix} 2a+1 \\ -2-a \\ 1-a \end{pmatrix} $$
To complete your projective scale, you also need to fix an origin, i.e. a point with coordinate “zero”, and a unit length, i.e. a point with coordinate “one”. Since all distances in your formula are measured from $G$, it makes sense to use that as the zero point. And since you could multiply all your lengths by a common scale factor without affecting the formula you stated, the choice of scale is irrelevant. Therefore we might as well choose $X$ as one. Note that this choice also fixes the orientation of length measurements along your line: positive is from $G$ in direction of $X$. We can then compute the two remaining coordinates, those of $Y$ and $Z$, using the cross ratio. In the following formula, square brackets denote determinants. The cross ratio of four collinear points in the plane can be computed as seen from some fifth point not on that line. I'll use $A$ for this purpose, both because it has simple coordinates and because, as stated above, the case of $l$ passing through $A$ has been omitted by the choice of coordinates for $l$.
$$
GY = \operatorname{cr}(F,G;X,Y)_A =
\frac{[AFX][AGY]}{[AFY][AGX]} =
\frac{\begin{vmatrix}
1 & 2a+1 & a \\
0 & -2-a & -1 \\
0 & 1-a & 0
\end{vmatrix}\cdot\begin{vmatrix}
1 & 1 & 1+a \\
0 & 1 & 0 \\
0 & 1 & 1
\end{vmatrix}}{\begin{vmatrix}
1 & 2a+1 & 1+a \\
0 & -2-a & 0 \\
0 & 1-a & 1
\end{vmatrix}\cdot\begin{vmatrix}
1 & 1 & a \\
0 & 1 & -1 \\
0 & 1 & 0
\end{vmatrix}}
= \frac{a-1}{a+2}
\\
GZ = \operatorname{cr}(F,G;X,Z)_A =
\frac{[AFX][AGZ]}{[AFZ][AGX]} =
\frac{\begin{vmatrix}
1 & 2a+1 & a \\
0 & -2-a & -1 \\
0 & 1-a & 0
\end{vmatrix}\cdot\begin{vmatrix}
1 & 1 & 0 \\
0 & 1 & 1+a \\
0 & 1 & a
\end{vmatrix}}{\begin{vmatrix}
1 & 2a+1 & 0 \\
0 & -2-a & 1+a \\
0 & 1-a & a
\end{vmatrix}\cdot\begin{vmatrix}
1 & 1 & a \\
0 & 1 & -1 \\
0 & 1 & 0
\end{vmatrix}}
= \frac{1-a}{1+2a}
$$
The thrid length, $GX$, is $1$ by the definition of the projective scale. So now you have the three lengths and plug them into your formula.
$$
\frac1{GX} + \frac1{GY} + \frac1{GZ} =
\frac{a-1}{a-1} + \frac{a+2}{a-1} - \frac{2a+1}{a-1} = 0
$$
As you will notice, the case of $a=1$ is problematic in this setup, since it would entail a division by zero. This corresponds to the situation where $l$ is parallel to $AB$. In that case, the point $X$ which we used as the “one” of the scale would coincide with the point $F$ at infinity, thus breaking the scale. A renaming of triangle corners will again take care of this special case. The cases $a=-2$ and $a=-\tfrac12$ correspond to the two other cases where $l$ is parallel to one of the edges. You will notice that the lengths would again entail divisions by zero, since the points of intersection are infinitely far away. But the reciprocal lengths are just fine and will be zero in those cases.
Best Answer
Too long for a comment ...
Let's try tightening-up your description. I'll note that we can eliminate the need to state (and repeat) "no sides of a triangle are parallel", and/or to introduce $X_1$, by simply identifying the triangle in question.
To begin, I believe this captures your set-up:
Now, you want to claim that each point lies on a separate edge of the triangle. As you indicate, this is a consequence of $\ell$ containing no vertices:
Your second paragraph discusses a point $X_1$, which we already have: it's $R$. So,
And then ... your third paragraph. It's not clear what's going on here. You want to "check if $C$ is on $c$" (although we have already declared that it is), but even so, your unsupported assertion about not being able to find an intersection with $\ell$ doesn't seem to have any bearing on the "check" your doing with line $c$. I'm not sure how to advise fixing this, because I seem to have fallen off of your train of thought.
Incidentally: When you get into these kinds of "intuitively obvious" arguments, you have to be really clear about your assumptions. It's especially important in this case, because your result is false in certain geometries.
Below is a picture of the Fano plane. It's a complete picture of the entire geometry, which consists of only seven points (represented by dots) and seven lines (represented by the segments and the circle). You can check that some fundamental geometric notions apply here: any two points lie on exactly one line; and any two lines meet at exactly one point. (There are no parallels here.) It's a perfectly good geometry ... but ... if we label the outer points $PQR$, the center point $O$, and the rest $A$, $B$, $C$, then "line $ABC$" contains exactly one point from each side of $\triangle PQR$.
Your response is likely to be "That's not what I'm talking about! I'm talking about good ol' regular geometry with infinitely many points and no silly circular lines!" And that's the point. You obviously (and very reasonably) want to rule out weird cases like the Fano plane ... because it probably (and very reasonably) never even occurred to you. But to do so, you have to be exceedingly careful not to argue by intuition: "There is no way how could we draw a line segment between two points on them (after they go through $A$ and $B$) and get intersection with $\ell$" isn't a valid argument to make, because there is a way on the Fano plane. You need to explain what it is about good ol' regular geometry that lets us know we aren't on the Fano plane.
If this nit-picking seems crazy, then welcome to the 19th century! It's around that time that mathematicians started realizing that they'd been making far too many intuitive assumptions about geometry for far too long, and they started establishing super-nit-picky foundations for the subject. Take a look, for instance, at David Hilbert's axioms for geometry; where Euclid thought just five postulates were sufficient to describe good ol' standard geometry, Hilbert realized there should be twenty to properly document our intuition.
(The good news is that the nit-picking only happens at the very-very fundamental level, with results such as yours that get to the very heart of what it means, for instance, for points to be collinear. Once we've essentially agreed on those foundations, we can happily move on without thinking too hard about them any more. After all, you don't see many geometry posts here at M.SE that start out saying "Assuming our geometry is governed by Hilbert's axioms, blah, blah, blah ..."; we just invoke the Pythagorean theorem or the Inscribed Angle Theorem or whatever and get about our business.)