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Playing with an ellipse I discovered the following properties and am now looking for a nice proof or references.

Let's consider an ellipse with foci $A$ and $B$ and let $C$ be a point on it. Let $l$ be the tangent to the ellipse passing through $C$. Let $C_a$ and $C_b$ be the circles with centers resp. A and B passing through $C$.

(P1) Prove that the other intersection point $F$ of the circles $C_a$ and $C_b$ lies on the ellipse.

Let $D = C_a \cap l$, $E = C_b \cap l$ and $H=DA \cap BE$

(P2) Prove that H lies on the ellipse

Let $I=C_a\cap HF$ and $J=C_b\cap HF$

(P3) Prove that the triples $(I, A, C)$ and $(J, B, C)$ are collinear.

Let $K=FH\cap l$, $L=DI\cap CJ$ and $M=CI\cap EJ$

(P4) Prove that $K$, $L$ and $M$ are collinear (is this some form of Pascal's theorem?)

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P1 follows from reflective symmetry across the major axis. – Blue Sep 11 '12 at 13:14
Good point. My chart needs simplification... – ivan Sep 11 '12 at 13:22
For P2: $H$ is specifically the reflection of $C$ about the center of the ellipse. A little angle-chasing and similar triangles will get you there. The key observation: the "mirror" property of ellipses guarantees that $\angle ACD = \angle BCE$. – Blue Sep 11 '12 at 14:14
For P3: What is $G$? – Blue Sep 11 '12 at 14:28
Okay, for P3: By my previous comments, it's clear that $C$, $F$, and $H$ are three corners of a rectangle with edges parallel to the axes of the ellipse. In particular $FH \parallel AB$. From this (and more angle and segment chasing), you can show $\triangle ABC \cong \triangle BAH \cong \triangle IHA \cong \triangle HJB \sim \triangle IJC$, so that $A$ and $B$ lie on $CI$ and $CJ$, respectively. – Blue Sep 11 '12 at 14:41

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