Prove that points $C,D,E,F,K,H$ lie on an ellipse $ABCDEFG$ is a regular heptagon. $H = AC \cap BD, K = AD \cap BF$ then show that $CDEFKH$ lie on an allipse.

Five points always define a conic, so I figured that is not hard to see that the conic through $FEDCH$ must be an ellipse due the configuration of the points. I don't have a clue about how to solve it, looks like some projective result which I'm not familiar with. I thought about the reverse of Pascal's theorem (Braikenridge-Maclaurin) but it was to no avail. May be related to Poncelet's theorem.
 A: Here is a proof using the reciprocal of Pascal's theorem:

Let $T$ be the mid point of $AB$. Then $E$ is a symmetry axis of the heptagon, so the lines $ET$, $AD$, $BF$ intersect in $K$.
Then in $\Delta KAB$ the angle bisectors $K$, $AHC$, $BG$ intersect in a point, denoted by $S$.
Consider now the hexagon
$$
EKFCHD\ .
$$
As also shown in the picture, the three relevant intersections
$EK\cap CH=S$, $KF\cap HD=B$, and the infinity point of the direction $DE\|FC$
are colinear. (The line $SB$ also passes through this infinity point, since it has the same direction $BSG\|CF\|DE$.)
The converse of Pascal theorem insures now the fact that the mentioned hexagon
has vertices on a conic. The conic is an ellipse $\mathcal E$, with one symmetry axis $s$ the common side bisector of $DE$ and $CF$, which contains $A$. It also passes through the reflections w.r.t. the axis $s$,

*

*$K'=AE\cap CG$ of $K=AD\cap BF$, and

*$H'=AF\cap GE$ of $H=AC\cap BD$.

$\square$

We are done, but let us...
Note: The tangent to $\mathcal E$ in the point $C$ is the heptagon side $BC$. And mirroring w.r.t. $s$ the tangent in $F$ is $GF$. To show this, one may consider again a hexagon like $DCCEKF$, and show that the line through $DC\cap EK$ and $S'=CE\cap DF$ also passes through $B$. (The point $AKD\cap CF$ is also on this line.)
A: You can assume without loss of generality that the heptagon is centered at the origin, and that its radius is $1$.
The first step is to generate the coordinates of the points
$E = (0, 1) $
Let $\theta = \dfrac{2 \pi}{7} $, then
$F = (- \sin \theta , \cos \theta ) $
$D = ( \sin \theta , \cos \theta ) $
$C = (\sin(2 \theta) , \cos(2 \theta) ) $
Diagonal $AD$ has a slope of
$ s_K = \dfrac{\cos \theta - \cos(3 \theta) }{ \sin \theta + \sin( 3 \theta) } $
Therefore,
$ K = ( 0,  \cos(3 \theta) + s_K \sin(3 \theta) )$
For point $H$, rotate the heptagon by $\theta$ clockwise, then the image of $H$ will be on the $y$ axis, and similar to the above
$H' = (0, \cos(3 \theta) + s_{H'} \sin(3 \theta) )$
where $s_{H'} = \dfrac{\cos(2\theta) - \cos(3 \theta) }{ \sin(2 \theta) + \sin( 3 \theta) }$
Finally,
$H = R_{\theta} H' $
where $ R_{\theta} = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $
At the final step, we have to verify that the following determinant is zero.
$\begin{vmatrix} x_1^2 && x_1 y_1 && y_1^2 && x_1 && y_1 && 1 \\
x_2^2 && x_2 y_2 && y_2^2 && x_2 && y_2 && 1 \\
x_3^2 && x_3 y_3 && y_3^2 && x_3 && y_3 && 1 \\
x_4^2 && x_4 y_4 && y_4^2 && x_4 && y_4 && 1 \\
x_5^2 && x_5 y_5 && y_5^2 && x_5 && y_5 && 1 \\
x_6^2 && x_6 y_6 && y_6^2 && x_6 && y_6 && 1 \end{vmatrix} $
where the $(x_i, y_i) , i = 1, 2, 3, 4, 5, 6 , $ are the coordinates of the six points.
I've written a small program to implement the above calculations, and the above determinant evaluated to $0$.  And this means the $6$ points lie on the same conic.
