A geometric property involving a cyclic quadrilateral and a conic

Question

Yesterday, while experimenting with GeoGebra, I discovered what seems to be a remarkable geometric property involving a cyclic quadrilateral and conic sections. However, I have not been able to prove it yet.

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The property

Let $ABCD$ be a cyclic quadrilateral, and let $G$ be any conic passing through the four vertices $A, B, C, D$. Let $P$ be an arbitrary point in the same plane.

From $P$, draw four perpendiculars to the sides of the quadrilateral:

\begin{aligned} L_1 &\perp AB,\\ L_2 &\perp BC,\\ L_3 &\perp CD,\\ L_4 &\perp DA. \end{aligned}

Each line $L_i$ intersects the conic $G$ at two points:

\begin{aligned} L_1 \cap G &= \{a, c'\},\\ L_2 \cap G &= \{b', d'\},\\ L_3 \cap G &= \{b, d\},\\ L_4 \cap G &= \{a', c\}. \end{aligned}

Now define the following intersection points:

\begin{aligned} W &= (ab) \cap (a'd'),\\ X &= (ab) \cap (b'c),\\ Y &= (b'c) \cap (c'd),\\ Z &= (c'd) \cap (a'd'). \end{aligned}

Claim: The quadrilateral $WXYZ$ is cyclic. enter image description here

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Questions

1. How can this property be proved? (I would prefer an Euclidean proof if possible.)

2. Is this property already known? If so, could you please provide a reference or link?

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Additional notes

1. I noticed that if the chosen conic $G$ coincides with the circumcircle of $ABCD$, then the chosen point $P$ and the centers of the two circles (the original and the constructed one) are collinear (see the purple line in the attached figure). enter image description here

2. A possibly related theorem states that:

If two conics intersect in four points, their axes of symmetry are perpendicular if and only if the four intersection points lie on a circle. A related discussion can be found here: On a problem concerning two conics (ResearchGate)

3. I have tried to generalize this theorem in several ways, and the configuration described above is the best formulation I have reached so far. I would be genuinely surprised if someone could extend it further.

4. The property still holds in degenerate cases — for instance, when two of the perpendiculars coincide, the corresponding connecting segment becomes a tangent to the conic.

5. I have a strong feeling that this theorem could have important implications and potential applications in various geometric constructions, though I have not yet explored them fully.

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Answer 1

1. How can this property be proved? (I would prefer an Euclidean proof if possible.)

I guess this is not quite the "Euclidean proof", yet it is a beautiful one.

Define $q_1=AD\cap a’c$, $q_2=BC\cap a’c$, $q_3=BC\cap b’d’$, $q_4=AD\cap b’d’$, and similarly, define $r_1,r_2,r_3,r_4$ given by intersecting lines $AB$, $CD$, $bd$, $ac’$ accordingly. Note that $q_1q_2q_3q_4$ and $r_1r_2r_3r_4$ are cyclic. With this in mind, we find the following generalization of the original question:

Generalized problem. Let $p_1,p_2,p_3,p_4,I,J$ be six points on a conic $C$. A conic $C_1$ intersects $p_1p_4$ in $q_2$ and $q_3$, and $p_2p_3$ in $q_1$ and $q_4$. Similarly, a conic $C_2$ intersects $p_1p_2$ in $r_3$, $r_4$, and $p_3p_4$ intersects $r_1$, $r_2$. A third conic $C_3$ intersects the lines $q_1q_4$, $q_2q_3$, $r_1r_4$, $r_2r_3$ in eight (8) pink points according to below diagram. Then the blue points in the diagram lie on a conic through $I$ and $J$.

Consequence. Picking $p_1p_2p_3p_4$=$ABCD$ and $I,J$ be the circle points (that is, $I=[1:i:0]$ and $J=[1:-i:0]$ in $\mathbb{P}^2$), we deduce the original problem.

Now, for the proof, we will use the following theorem due to David Eisenbud, Mark Green, and Joe Harris in the paper "CAYLEY-BACHARACH THEOREMS AND CONJECTURES":

Theorem. Main theorem. Let $X_1,X_2\subset \mathbb{P}^2$ be plane curves of degree $d$ and $e$, respectively, intersecting in $d\cdot e$ points $\Gamma=X_1\cap X_2=\{p_1,\dots,p_{de}\}$, and suppose that $\Gamma$ is the disjoint union of subsets $\Gamma'$ and $\Gamma''$. Let $s=d+e-3$. If $k\leq s$ is a nonnegative integer, then the dimension of the vector space of homogeneous polynomials of degree $k$ vanishing on $\Gamma'$ (modulo those containing all of $\Gamma$) is equal to the failure of $\Gamma''$ to impose independent conditions on homogeneous polynomials of degree $s-k$.

Proof. See the article.

Proof of generalized problem. First, redefine the blue point in the top right as the intersection of the conic passing through the three (3) other blue points and $I,J$ with the red line. We are asked to prove that the blue point lies on the blue line.

enter image description here

Let $X_1$ be the union of the six (6) red lines and the purple line, so a curve of degree $d=1\cdot 7$. Let $X_2$ be the union of the five (5) conics, so a curve of degree $e=2\cdot5=10$. So $\vert X_1\cap X_2\vert=d\cdot e=70$. Lets partition $\Gamma=X_1\cap X_2$ as follows: $$\Gamma’’=\{26 \text{ marked points in the picture}\}\cup \{p_1,p_2,p_3,p_4\}\cup\{3 \times I\} \cup\{3 \times J\} $$ Note that $p_i$ appears twice and $I$, $J$ appear four (4) times each in the intersection $X_1\cap X_2$. We let $\Gamma’$ be the other 70ドル-(26+4+3+3)=34$ points. $s=d+e-3=7+10-3=14$. Pick $k=7\leq s=14$. Then, since the vector space of degree $k=7$ forms (i.e., plane curves of degree 7ドル$) is of dimension $\binom{2+7}{2}-1=36-1=35>34=\vert\Gamma’\vert$, we see that the dimension of degree $k=7$ forms that vanish on $\Gamma’$ is non-zero. The theorem then states that $\Gamma’’$ fails to impose independent conditions on forms of degree $s-k=14-7=7$. That is, any degree $s-k=7$ curve passing through 35ドル$ points of $\Gamma’’$ must pass through the last point. In particular, we may conclude by considering the curve defined as the union of the six (6) blue lines and the purple line.