Require a Proof for Intelligenti pauca's method to Compute an Ellipse

Question

In smls's post "Find the properties of an ellipse from 5 points in 3D space" (Find the properties of an ellipse from 5 points in 3D space), Intelligenti pauca presented a pure geometrical answer. I tested that method and found it good. But I cannot prove the process of finding the tangent A-TA. Anyone can show me the proof?

Answer 1

That trick is based on Pascal's Theorem:

If six arbitrary points are chosen on a conic (which may be an ellipse, parabola or hyperbola) and joined by line segments in any order to form a hexagon, then the three pairs of opposite sides of the hexagon (extended if necessary) meet at three points which lie on a straight line, called the Pascal line of the hexagon.

You can see the theorem at work in the figure below: hexagon $A'ABEDC$ is inscribed in an ellipse and its three pairs of opposite sides (having the same colour in the figure) meet at points $F$ (intersection of $AB$ and $CD$), $G$ (intersection of $A'C$ and $BE$) and $H$ (intersection of $A'A$ and $DE$), which lie then on the same line.

Suppose now you let $A'$ approach $A$ closer and closer: in the limit $A'\to A$ line $AA'$ becomes the line tangent to the ellipse at $A$ (see second figure).

This gives a method to construct the tangent at $A$ to a conic passing through points $ABCDE$: it is the line through $A$ and $H$, the latter being the intersection point of lines $FG$ and $DE$. Points $F$ and $G$ are constructed as explained above but with $A'$ replaced by $A$: $F$ is the intersection of $AB$ and $CD$, $G$ is the intersection of $AC$ and $BE$.