Below is a proof of a slight generalization of "Blue's version" of the problem. The slight generalization is that $\overline{PQ}$ is merely assumed to be parallel to $\overline{BC}$; the conclusion is that $D$, $E$, and $Z$ in the figure are collinear.
To recap, we are assuming that:
- $\triangle ABC$ is a triangle with an altitude $\overline{AD}$
- $X = D_b$, where "$D_b$" denotes the projection of $D$ onto $\overline {AB}$
- $Y=D_c$, where "$D_c$" denotes the projection of $D$ onto $\overline{AC}$
- $\overleftrightarrow{PQ}$ is parallel to $\overline{BC}$, with $P$ on $\overline{AX}$ and $Q$ on $\overline{AY}$
- $Z$ is the intersection of $\overline{PY}$ and $\overline{QX}$
Draw the line through $P$ perpendicular to $\overline{AB}$ and the line through $Q$ perpendicular to $\overline{AC}$, and let $E$ be the intersection of those two lines. (In other words, $E_b = P$ and $E_c = Q$.)
The points on line $\overleftrightarrow{DE}$ are the points which can be written in the form $(1-\lambda)D + \lambda E$ (with $\lambda \in \mathbb{R}$). For all $\lambda$, the point $K = (1-\lambda)D + \lambda E$ satisfies
\begin{align*}
K_b &= (1-\lambda)D_b + \lambda E_b = (1-\lambda)X + \lambda P \\
K_c &= (1-\lambda)D_c + \lambda E_c = (1-\lambda)Y + \lambda Q.
\end{align*}
The "converse" is also true, in the following sense: For any point $K$ (not necessarily on $\overleftrightarrow{DE}$), let $\lambda_b(K)$ be the unique number $\lambda_b$ such that $K_b = (1-\lambda_b)X + \lambda_b P$, and let $\lambda_c(K)$ be the unique number $\lambda_c$ such that $K_c = (1-\lambda_c)Y + \lambda_c Q$; then $\overleftrightarrow{DE}$ is the locus of points $K$ for which $\lambda_b(K) = \lambda_c(K)$.
(You could think of $(\lambda_b, \lambda_c)$ as a bizarre oblique coordinate system for the plane.)
Oh yeah, so what were we trying to prove again?
Claim: $Z$ lies on $\overleftrightarrow{DE}$.
Special case (corollary): In the case where $P$ is midpoint of $\overline{AB}$ and $Q$ is the midpoint of $\overline{AC}$, the point $E$ is in fact the circumcenter $O$. Hence the conclusion is taht $Z$ must lie on line $\overleftrightarrow{DO}$.
Proof of the claim:
Use the right angles at $X$, $D$, and $Y$ to show that $\angle ADX = \angle B = \beta$ and $\angle ADY = \angle C = \gamma$.
Since $\angle AXD$ and $\angle AYD$ add up to $180^{\circ}$, the quadrilateral $AXDY$ is a cyclic quadrilateral. This implies that $\angle AYX = \angle ADX$ and $\angle AXY = ADY$.
We now know the angles at all four vertices of the quadrilateral $XYQP$: $\angle X = \gamma$ and $\angle Y = \beta$ from the above, and $\angle Q = 180^{\circ}-\gamma$ and $\angle P = 180^{\circ}-\beta$ since $\overline{PQ}$ is parallel to $\overline{BC}$.
These angles show that $XYQP$ is a cyclic quadrilateral. In particular, this means that $\triangle PZX$ and $\triangle QZY$ are similar. The similar triangles imply that $\lambda_b(Z) = \lambda_c(Z)$.
From the earlier discussion, we conclude that $Z$ lies on $\overleftrightarrow{DE}$.
(Remark: The center of the circle passing through $X$, $Y$, $Q$, and $P$ also lies on $\overleftrightarrow{DE}$, since it has $\lambda_b = \lambda_c = 1/2$.)
But wait, what does it have to do with the Euler line?
Uhhhh... nothing really, as far as I can tell. In my mind the Euler line has to do with that one neat trick of using a dilation by a factor of $-2$ centered at the centroid. This problem doesn't seem to involve that idea at all. In my proof here, the outline was that $\overleftrightarrow{DE}$ is characterized by "$\lambda_b = \lambda_c$", and $Z$ has that property. Of course, I don't know what proof of the existence of the Euler line is the one you have in mind, but I personally don't see any helpful analogy.
