Given a pair of inversely similar triangles ABC and AB'C', as shown in the graph, CB and C'B' intersect at F. X, Y, Z are the midpoints of BB', CC', and AF respectively.
Prove that X, Y, Z are co-linear.
This is a very elegant result, and somewhat alike Newton's line.
All proofs welcome, and a geometrical solution would be very nice! 
On
Here's a coordinate approach.
Position $A$ at the origin, and let the positive $x$-axis bisect $\angle BAB'$. Then we can assign coordinates $$\begin{align} B := 2 b\,(\,\cos\beta, \sin\beta\,) &\qquad B' := 2 b \lambda\,(\cos(-\beta), \sin(-\beta)\,) \,=\, 2 b \lambda\,(\cos\beta, -\sin\beta\,) \\[4pt] C := 2 c\,(\,\cos\gamma, \sin\gamma\,) &\qquad C' := 2 c\lambda\,(\cos(-\gamma), \sin(-\gamma)\,) \,=\, 2 c\lambda\,(\cos\gamma, -\sin\gamma\,) \end{align}$$ for some $b$, $c$, $\beta$, $\gamma$, $\lambda$.
If $b\sin\beta=c\sin\gamma$, then lines $BC$ and $B'C'$ (and line $XY$) are parallel to the $x$-axis, making $F$ the point-at-infinity of these lines.
If $b\cos\beta=c\cos\gamma$, then lines $BC$ and $B'C'$ coincide with a perpendicular to the $x$-axis; that perpendicular contains $F$, $X$, $Y$.
Otherwise, barring other degeneracies, we have $$F = bc\sin(\beta-\gamma)\,\left(\frac{\lambda+1}{b \sin\beta - c \sin\gamma}, \frac{\lambda-1}{b \cos\beta - c \cos\gamma}\right) \qquad $$ $$X = \left(\,(\lambda+1) b \cos\beta, -(\lambda-1) b \sin\beta\,\right) \qquad Y = \left(\,(\lambda+1) c \cos\gamma, -(\lambda-1) c \sin\gamma\,\right) \qquad Z := \frac12F$$ Without too much trouble, we can determine that the equation of line $XY$ is given by $$ x\,(\lambda-1)\,(b \sin\beta - c \sin\gamma) \;+\; y\,(\lambda+1)\,(b \cos\beta - c \cos\gamma) \;=\; (\lambda^2-1) \,b c \sin(\beta-\gamma)$$ Clearly, $Z$ satisfies this equation, so that $X$, $Y$, $Z$ are collinear. $\square$
The proof described below assumes that the pair of inversely similar triangles $ABC$ and $ADE$ are so positioned that their sides $BD$ and $DE$ are not parallel. This assumption is made to guarantee the existence of point $F$, without which the configuration proposed by OP is not possible. Before attempting to tackle the question OP posed in the problem statement, we need to prove two lemmata.
$\underline{\bf{Lemma\space 1}}:$
As shown in $\mathrm{Fig.\space 1}$, a pair of inversely similar triangles $ABC$ and $ADE$ are drawn so that they share the vertex $A$. The side $CA$ of $\triangle ABC$ is extended to meet the side opposite the common vertex of $\triangle ADE$ at $Q$. Similarly, the side $EA$ of $\triangle ADE$ is extended to meet the side opposite the common vertex of $\triangle ABC$ at $P$. Then we have, $\dfrac{BP}{PC}=\dfrac{DQ}{QE}$.
$\bf{Proof\space of\space Lemma\space 1}:$
Let $DE=k\times BC$. Since they are opposite angles, $\measuredangle CAP = \measuredangle EAQ$. Since $\triangle ABC$ and $\triangle ADE$ are inversely similar, $\measuredangle BCA = \measuredangle AED$. Since two of the angles of $\triangle APC$ are equal to the corresponding angles of $\triangle AQE$, this pair of triangles are inversely similar. Therefore, we have, $$QE=k\times PC, \tag{1}$$ and we shall write, $$DE-QE=k\times\left(BC-PC\right)\qquad\rightarrow\qquad DQ=k\times BP. \tag{2}$$
We can divide (2) by (1) to obtain $\dfrac{DQ}{QE}=\dfrac{BP}{PC}$.
$\underline{\bf{Lemma\space 2}}:$
As shown in $\mathrm{Fig.\space 2}$, $BCED$ is an arbitrary quadrilateral. If points $P$ and $Q$ are located on the sides $BC$ and $DE$ respectively such that $BP\space :\space PC=DQ\space :\space QE$, then the midpoints of the segments $BD$, $PQ$, and $CE$ are collinear.
$\bf{Proof\space of\space Lemma\space 2}:$
Let $X$, $W$, and $Y$ be the midpoints of the segments $BD$, $PQ$, and $CE$ respectively. We draw the segments $PD$ and $PE$ and mark their respective midpoints as $M$ and $N$. We also need to draw three segment to join $X$ to $M$, $M$ to $N$, and $N$ to $Y$. Please note that $W$ lies on $MN$. Denote $BP=a$, $PC=b$, and $BP\space :\space DQ=1\space :\space k$.
By applying the midpoint theorem to $\triangle PDB$ and $\triangle CEP$, we can show that $XM$ is parallel to $NY$. Since the points $M$, $W$, and $N$ are collinear, we have, $$\measuredangle XMW=\measuredangle WNY \tag{3}$$
When the same theorem is applied to $\triangle PDB $, $\triangle DPQ$, $\triangle QPE$, and $\triangle CEP$, we get, $$XM=\dfrac{a}{2}, \qquad MW=\dfrac{ka}{2}, \qquad NW=\dfrac{kb}{2}, \quad\space\text{and}\quad\space NY=\dfrac{b}{2}.$$
This means that, $$XM\space :\space MW=NY\space :\space NW. \tag{4}$$
Using (3) and (4), we can state that the pair of triangles $XMW$ and $YNW$ are similar.
$$\therefore\quad \measuredangle MWX=\measuredangle NWY. \tag{5}$$
We also know that the three points $M$, $W$, and $N$ are collinear. As a consequence of this and (5), the three points $X$, $W$ and $Y$ are also collinear.
$\bf{\text{Proof of Collinearity of Points $X$, $Y$, and $Z$}}:$
In $\mathrm{Fig.\space 3}$, we reproduce OP’s sketch after adding a few segments. We have extended $EA$ and $CA$ until they meet $BC$ and $DE$ at $P$ and $Q$ respectively. The segment $PQ$ is also drawn and $W$ is its midpoint. As we have mentioned at the beginning of this answer, we assume that the two lines $BC$ and $DE$ are not parallel.
It follows from Lemma 1 that $BP\space :\space PC=DQ\space :\space QE$. Therefore, we can apply Lemma 2 to the quadrilateral $BCED$ to confirm that the three points $Y$, $W$, and $X$ are collinear. Call the line on which these three points sit $L_{\large{1}}$.
The Newton-Gauss theorem asserts that the midpoints of the three diagonals of any complete quadrilateral are collinear and the line on which these points are located is known as the Newton-Gauss line. Now, consider the complete quadrilateral $FCBADE$, the four sides of which are $CBF$, $EDF$, $CAQ$, and $EAP$, Please note that no two of the mentioned four lines are parallel and no three are concurrent. Its three diagonals are $CE$, $PQ$, and $FA$. Their mid points are $Y$, $W$, and $Z$ respectively and they are collinear. For brevity, let us call this line $L_{\large{2}}$.
Since both $L_{\large{1}}$ and $L_{\large{2}}$ pass through $Y$ and $W$, these two lines are one and the same. This means that point $X$, which lies on line $L_{\large{1}}$, and point $Z$, which lies on line $L_{\large{2}}$, are located on the line $YW$. Therefore, $X$, $Y$, and $Z$ are collinear.