Proving Archimedes PROP VII-12: In an ellipse, the sum of the squares on conjugate diameters equals the sum of the squares on its axes I was asked the following question:

"Prove analytically Proposition VII-12 that in any ellipse the sum of the squares on any two of its conjugate diameters is equal to the sum of the squares on it's two axes."


Above is my drawing of conjugate diameters. I know that the diameters are not perpendicular to each other - and this is clear from the picture. I would think I can use the pythagorean theorem in some way to prove this because what the proposition is essentially saying is that $HJ^2+ID^2=FG^2+AB^2$ - which has the look of the Pythagorean theorem.
Can anyone help me take the next step?
 A: One approach is to take the slope of $ED$ equal to $k$, making the point $D$ be $(x, kx)$, use calculus to compute the slope $K$ of $EH$ and then just compute the sum of squares. Namely, if the equation of the ellipse $ax^2+by^2=1$ then we have $x^2(a+bk^2)=1$ and $ED^2=\frac{1+k^2}{a+bk^2}$. The tangent at $D$ is given by implicit differentiation as $2axdx+2bydy=0$, so slope is $dy/dx=-a/(bk)$. Now we can compute $EH^2$ by the same formula just with the new slope instead of $k$. Then just algebraically check the sum is independent of $k$ and is indeed equal to what one would get when $k=0$ to which this computation is inaplicable.
Here is another approach (without derivatives): the ellipse with major axis of length $a$ and minor of length $b$ (NOT the same $a$ and $b$ as in the first approach) is a the image of the circle $x^2+y^2=1$ under the stretching map, and  so inherits a parameterization $(a\cos t, b \sin t)$ (note that $t$ is the angle along the circle, but NOT along the ellipse, so the conjugate diameters are not orthogonal). The stretching map takes tangents to tangents and since it is linear it preserves parallelism, so takes conjugate diameters on the circle to conjugate diameters on the ellipse. Thus to get, say, $H$ from $D$ one only needs to increase $t$ by $\pi/2$, so if $D$ has coordinates $(a\cos t, b \sin t)$ then $H$ has coordinates $(a\cos (t+\pi/2), b \sin (t+\pi/2))$. The sum of square lengths is evidently constant at $a^2+b^2$.
The linear algebra version is like this: let $A:\mathbb{R}^2\to \mathbb{R}^2$ be any linear map (in this problem, $A$ is any map taking the circle to the ellipse, say the stretching map) and $v_1$ $v_2$  an orthonormal basis for its domain. Then $|Av_1|^2+|Av_2|^2$ (which in this problem is the sum of squared lengths of the "conjugate axis")  is independent of the choice of the basis. Indeed, let $V$ be the rotation matrix with columns $v_1$ and $v_2$. Then
$$|Av_1|^2+|Av_2|^2=<Av_1, Av_1>+<Av_2, Av_2>=<v_1, A^TAv_1>+<v_2, A^TAv_2>=$$
$$v_1^T(A^TA)v_1+v_2^T(A^TA)v_2= $$
$$tr(V^TA^TAV)=tr(V^{-1}A^TAV)=tr(A^TA)$$
is constant.
