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".)
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:
Suppose line $\ell$ contains exactly three distinct, non-vertex points of $\triangle PQR$, and call those points $A$, $B$, $C$.
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:
No two of these points lie on one edge of the triangle (otherwise, $\ell$ would coincide with that edge and contain the vertices at its endpoints, but $\ell$ contains no vertices). We'll say that $A$ lies on the edge opposite $P$; $B$ lies on the edge opposite $Q$, and $C$ lies on the edge opposite $R$.
Your second paragraph discusses a point $X_1$, which we already have: it's $R$. So,
Since $C$ doesn't lie on $\overleftrightarrow{PR}$ or $\overleftrightarrow{QR}$, it certainly doesn't lie on the rays $\overrightarrow{RA}$ or $\overrightarrow{RB}$ within those lines.
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.)
Best Answer
Triangles $ EMB $ and $ CDE $ are similiar, so yes, the last relation you mentioned holds.