TITLE: When is the Euler line parallel with a triangle's side?
QUESTION [5 upvotes]: When is the Euler line parallel with a triangle's side?
I have found that a triangle with angles $45^\circ$ and $\arctan2$ is a case.
Is there any other case?
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REPLY [4 votes]: Here's an approach that's more satisfying than my previous attempt.

Let $\triangle ABC$ have midpoints $D$, $E$, $F$ opposite $A$, $B$, $C$. Let it have centroid $K$, circumcenter $P$, and orthocenter $Q$. Let $G$ be the foot of the perpendicular from $C$, and let $M$ be the midpoint of $\overline{CG}$.

Suppose the Euler line (through $K$, $P$, $Q$) is parallel to $\overline{AB}$. Then, since we "know" $K$ trisects median $\overline{CF}$, we can get the following chain of ratios:
$$\begin{align}
3 &= \frac{|CF|}{|KF|}=\frac{|CM|}{|QM|}=\frac{|\triangle CDE|}{|\triangle PDE|}=\frac{\frac12|CD||CE|\sin DCE}{\frac12|PD||PE|\sin DPE}=\frac{|CD|}{|PD|}\cdot\frac{|CE|}{|PE|} \\[4pt]
&=\tan A\tan B
\end{align}$$
Here, we used the fact that $\angle DPE=180^\circ-\angle DCE$ (so that the sines cancel). Also, by the Inscribed Angle Theorem, $\angle BPC=2\angle A$, so that $\angle DPC =\frac12 \angle BPC = \angle A$ (and likewise, $\angle EPC=\angle B$). $\square$