While working on conic curve problems, I have observed certain similarities.
Let $\Gamma$ be a conic curve in the plane, and let $A$ be a moving point on $\Gamma$. Fix two points $B$ and $C$ in the plane. The lines $AB$ and $AC$ intersect $\Gamma$ at points $D$ and $E$ respectively. My goal is to demonstrate that $DE$ is always a tangent to another fixed conic curve.
I attempted to perform a projective transformation such that $BC$ is projected onto an infinite line. Subsequently, an affine transformation is applied so that $\Gamma$ becomes either a circle, a parabola, or a hyperbola.
If the outcome of the affine transformation results in a circle, proving the theorem is straightforward, given the fixed angles formed by the lines $AB$ and $AC$ . However, proving the remaining cases through algebraic manipulation proves to be challenging.
If $BC$ does not intersect with $\Gamma$, the entire process is simplified. In this case, we can first project $\Gamma$ onto a circle and then project $BC$ onto an infinite line, making the entire process more manageable.
Please let me know if a geometric background is required for the problem. Any form of assistance would be appreciated!
UPD: An example image of the description above are as follows, where $DE$ defined above is marked as "$EG$":
This is an algebraic approach. Let's assume your conic $\Gamma$ is given by a symmetric matrix, for which I'll re-use the symbol $\Gamma$. $A^T\cdot\Gamma\cdot A=0$ expresses that $A$ lies on $\Gamma$.
Since $D$ is a point on the line $AB$ and distinct from $A$ can be described as a linear combination $\lambda A+B$. It also lies on $\Gamma$.
$$0=D^T\cdot\Gamma\cdot D= (\lambda A+B)\cdot\Gamma\cdot(\lambda A+B)= \lambda^2\underbrace{(A^T\cdot\Gamma\cdot A)}_{=0}+ 2\lambda(A^T\cdot\Gamma\cdot B)+ (B^T\cdot\Gamma\cdot B)$$
But since $A$ lies on $\Gamma$ the quadratic term is zero. So you can solve the remaining linear equation for $\lambda$:
$$\lambda=-\frac12\frac{B^T\cdot\Gamma\cdot B}{A^T\cdot\Gamma\cdot B}$$
Choosing a different representative you can write
$$D=(B^T\cdot\Gamma\cdot B)A-2(A^T\cdot\Gamma\cdot B)B$$
Likewise for $E$:
$$E=(C^T\cdot\Gamma\cdot C)A-2(A^T\cdot\Gamma\cdot C)C$$
The line joining $D$ to $E$ is the cross product $h=D\times E$. Computed in this way, the vector $h$ is quadratic in each of $A,B,C$ because the degrees from $D$ (namely $A^1B^2C^0$) and $E$ (namely $A^1B^0C^2$) add up. But that vector carries a common factor $\det(A, B, C)$. The geometric interpretation behind that factor is that if $A, B, C$ are collinear, then $D\sim E$ (they are the same point but not necessarily the same homogeneous vector) and the line joining them becomes undefined, represented as the null vector. Canceling that common factor makes $h$ linear in each of $A,B,C$. This corresponds to removing a removable singularity. If you fix $B$ and $C$, then $h$ is linear in $A$, so you can see that as some matrix times vector $A$. And if you feed the points of a conic to a matrix, you get again a conic. Even if you interpret that second conic is in your setup as dual conic, a conic characterized by tangent lines not incident points. But algebraically that's the same structure. Q.e.d.
I'm not entirely happy with this proof. I can argue well that the factor $\det(A, B, C)$ has to be present in $h$, but I can not demonstrate it by looking at the formula alone. Nor can I show you what the vector looks like if you cancel that factor. At least not without breaking things down to the coordinate level. I did some experiment using the unit circle as the initial conic:
$$ A=\begin{pmatrix}t^2-u^2\\2tu\\t^2+u^2\end{pmatrix} \qquad \Gamma=\begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix} $$
Since any non-degenerate conic (with real points on it) can be transformed to the unit circle using a suitable projective transformation, and since your statement is invariant under projective transformation, a proof using the unit circle as my conic is still without loss of generality. Using that proof I could observe the factor $\det(A, B, C)$ and that the coordinates of $h$ are linear in those of $A$ (and in this case quadratic in $(t,u)$). But demonstrating this using the general formula instead of a concrete example would have been nicer.
I've also tried to experiment a bit with Cinderella to get a feeling for your configuration. I found that it might be beneficial to also illustrate the dual version of this problem:
Start with a conic, in this case the dark green circle, and two lines $b$ and $c$. Pick a line $a$ tangent to the circle. From the point where $a$ and $b$ intersect (labeled $B$ above) draw the other tangent to the original conic, labeled $d$. Likewise from the point where $a$ and $c$ intersect (labeled $C$ above), draw the other tangent $e$. Now observe the point $F$ which is the intersection of $d$ and $e$. The locus of $F$ as $A$ moves on the initial conic is again a conic, drawn in red. And the locus of a point is something that e.g. Cinderella is happy to construct, and the resulting image is less cluttered than all the tangent lines in the original post.
One observation you can get from this construction is that the original and the new conic touch in two points, $K$ and $L$. The tangents from point $Q$, which is the intersection of $b$ and $c$, also go through these points, as does the polar of $Q$. In your primal formulation, the conics touch as well. Your line $BC$ (which corresponds to my point $Q$) intersects the conic $\Gamma$ in these two touching points. If I move my point $Q$ into the interior of the initial conic, or you move the points $B$ and $C$ so that the line connecting them doesn't intersect the conic, then the touching points have complex coordinates but the algebra for them should still work the same.
Another interesting scenario is if you keep using a circle as the initial (green) conic, and then move the point $Q$ to the center of that circle. The resulting (red) conic is a concentric but larger circle, and it appears that the ratio between the two circle radii might be the cosine of the angle between line $b$ and line $c$.
Currently I don't know how to use the touching points or that concentric configuration to help answer your question in a more elegant or intuitive way. But perhaps this alternate perspective can help someone else find a different answer.