Let's imagine that we have an ellipse described by the known equation $v^TAv=0$, (Link_1) where $v=[x \ y \ 1]^T$
(it can be a skew one in a general case).
Then we have all possible parallel lines - for example horizontal lines on the $OXY$ plane: $y=b$. Some of these lines don't cross ellipse, two are tangent to ellipse at points $C_1$ and $C_2$, other cross it in two points $B_1$ and $B_2$.
Question:
Is there a relatively easy proof that midpoints of all segments $B_1B_2$ (preferable way to start from the mentioned above general equation) construct linear segment $C_1C_2$ and the midpoint of this segment $C_1C_2$ is the center of ellipse ?
Yes. Well...relatively easy if you're willing to allow me to use Sylvester's law of inertia, which says that every symmetric matrix over the reals is diagonalizable.
The matrix $A$ can be replaced by $(A + A^t)/2$ without changing the ellipse, so we may assume $A$ is symmetric.
Write $A = Q^t D Q$ for some orthogonal matrix $Q$ and diagonal matrix $D$. Under the transformation $v \mapsto \sqrt{D}Qv$, the ellipse $A$ gets mapped to the unit circle. (The matrix $D$ has a square root because your quadratic form determines an ellipse, i.e., both eigenvalues are positive.)
Under that same transformation, which involves only rotation (by $Q$) and scaling along the axes (by $\sqrt{D}$), the set of horizontal chords of the ellipse is transformed into a set of parallel chords of the circle, and the midpoint operation is preserved.
In the new coordinate system, the midpoints of a set of parallel chords of a circle form a segment that's a diameter of the circle... and we're done: we transform that segment back to get a straight segment in the original space, and its midpoint is still the center of the ellipse.
(To make all this work really cleanly, it helps to begin by translating the ellipse to have its center at the origin.)
A post-comment note: Suppose that $T$ is linear. I claim that for a segment $AB$, the midpoint of $AB$ transforms, under $T$, to the midpoint of the segment from $T(A)$ to $T(B)$.
Proof: The midpoint $M$ is just $0.5A + 0.5B$. When we apply $T$, we get \begin{align} T(M) &= T(0.5A + 0.5B) \\ &= T(0.5A) + T(0.5B) \\ &= 0.5 T(A) + 09.5 T(B) \end{align} which is the midpoint of the segment between $T(A)$ and $T(B)$.