Question about 'Archimedes parabola' In the Wikipedia article on lever mechanics from the Archimedes codex where 

"The first proposition states:
The area of the triangle ABC is exactly three times the area bounded by the parabola and the secant line AB."


The wikipedia proof ends prematurely in my view:

"In other words, it suffices to show that $EF:GD = EH :JD$. But that is a routine consequence of the equation of the parabola. Q.E.D."

I cannot see this is obvious - in fact it seems counterintuitive since the parabola equation is quadratic while the other dimensions are linear.
Please could you see if there is a explanation for the Q.E.D bit.

An analytic proof. Let $y=ax^2$ be the equation of a generic parabola, and 
$A=(x_1,ax_1^2)$, $B=(x_2,ax_2^2)$ any two points on it. The equation of tangent $BC$ is then $y-ax_2^2=2ax_2(x-x_2)$ and $C=(x_1,2ax_1x_2-ax_2^2)$. Point $D$ is the midpoint of $AC$, thus: $D=(x_1,ax_1x_2+a(x_1^2-x_2^2)/2)$.
Let now $x$ be the abscissa of $E$. We have immediately:
\begin{align}
& E=(x, ax(x_1+x_2)-ax_1x_2),\quad
F=(x, ax^2),\quad
H=(x, 2axx_2-ax_2^2), \\[8pt]
& G=\left(x, {1\over2}ax(x_1+3x_2)-{1\over2}ax_2(x_1+x_2)\right).
\end{align}
We get then:
$$
{EF\over EH}={ax^2-ax(x_1+x_2)+ax_1x_2\over2axx_2-ax_2^2-ax(x_1+x_2)+ax_1x_2}
={a(x-x_2)(x-x_1)\over a(x-x_2)(x_2-x_1)}={x-x_1\over x_2-x_1}
={GD\over BD},
$$
but this is the same as $EF:GD=EH:JD$, because $JD=BD$.
