Ruler and compass construction of the axis of an ellipse related to a triangle Let $ABC$ be a triangle. Consider all the six points defined as the opposite of a vertex respect to another.
It is not hard to prove (with affinity) that these 6 points all lie on an ellipse.
Looking on it on geogebra it also seems that the center of this elliplse is the centroid of $ABC$.
However I cannot find a way to construct the axis of this elliplse using just ruler and compass and I was wondering if maybe there is a way of doing it.
For some cases it is easy, for example if $ABC$ is isosceles with base $AB$ then one of the axis is just the perpendicular bisector of $AB$ and the other one would be the parallel to $AB$ throught the centroid, but for a general triangle this cannot apply.
 A: From your triangle $ABC$ you can obtain an isosceles triangle $ABD$ by applying a shear which leaves in place the points of line $AB$. And from $ABD$ you can get an equilateral triangle $ABE$ by a dilation along a direction perpendicular to $AB$.
In the case of $ABE$, the six points you described belong of course to a circle, having its centre at the centroid $L$ of $ABE$ and passing through points $A'$, $B'$, which are the symmetric of $A$ about $B$ and vice versa. A simple computation gives then $r=\sqrt{7/3}AB$ as the radius of the circle, and $LS/LE=\sqrt7$ (see figure below).
In the case of isosceles triangle $ABD$, that circle is dilated into an ellipse, centred at the centroid $K$ of $ABD$. One of its semi-axes is parallel to $AB$ and has length $r$, while the other is $PK$ which can be found from $PK/DK=LS/LE=\sqrt7$.
In the case of triangle $ABC$, the ellipse of $ABD$ gets sheared to another ellipse, centred at the centroid $G$ of $ABC$. The semi-axes of the old ellipse are sheared to two conjugate semidiameters: $GR=r$ parallel to $AB$ and $GQ$ such that $GQ/GC=PK/DK=\sqrt7$.
Once you have constructed two conjugate semidiameters, the axes of the ellipse can be constructed as explained here.

