One of many properties of an ellipse is this:
the locus of midpoints of parallel chords is a straight line.
A proof I present below is more or less brute-force. I am looking for a purely geometric proof, using perhaps the very definition of the ellipse as being the locus of points whose sum of distances from two fixed points (the foci) is constant, or perhaps some other geometric property. The question has appeared before, here. The answer given there, however, seems to me a bit vague.
The red dots are midpoints of parallel chords. The claim is that all other midpoints of additional parallel chords will lie on the blue dashed line. To prove this, suppose the ellipse is given in its canonical form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Suppose also that a fixed slope has been chosen for the parallel chords, so that the equation of the chord passing through the origin in the picture is $y=\beta x$ from some fixed $\beta >0$. Given any chord parallel to $y=\beta x$, let $t$ be such that $y=\beta(x+t)$. This $t$ serves as a parameter of the family of parallel chords, designating the point where the chord meets the $x$-axis. Each such chord, save the tangent ones, meets the ellipse in exactly two points, which are solutions of the equations $$y=\beta(x+t),\quad\hbox{and}\quad \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ Substituting for $y$, say, we get $$\frac{x^2}{a^2}+\frac{\beta^2(x+t)^2}{b^2}=1$$ that is, $$x^2(\beta^2a^2+b^2)+x(2a^2\beta^2t)+a^2(\beta^2t^2-b^2)=0$$ The $x$-coordinate of the midpoint of the chord containing the two solutions is of course their average, which is the sum of the roots of the quadratic above, divided by two. But we know how to add up the solutions of a quadratic. Hence the $x$ coordinate of the midpoint is $$-\left(\frac{a^2\beta^2}{\beta^2a^2+b^2}\right)t$$ Let $c_1$ denote the number in the brackets. It depends on the ellipse, and on the initial slope $\beta$. Thus the midpoint's $x$-coordinate has the form $-c_1t$. Since our midpoint belongs also to the line $y=\beta(x+t)$, we get that the corresponding $y$-coordinate is given by $$y=\beta(x+t)=\beta(-c_1t+t)=\beta(1-c_1)t=c_2t$$ It follows that all red midpoints belong to the line $\{t(-c_1,c_2):t\in\mathbb{R}\}$. Q.E.D.
P.S It is a proof, but it is not geometric. I bet Newton would not have proved it this way.
