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.
 A: The proportion Archimedes needs here rests on $Props. 4$ and $5$ of his Quadrature of the Parabola. In the posted figure, bisect $AB$ at $K$ and draw $KP$ parallel to $AC$, cutting the parabola at $M$, and from $F$ draw $FL$ parallel to $AB$, meeting $JB$ at $N$. $Prop. 4$ proves that $$\frac{BK}{KE}=\frac{EG}{FG}$$For since $MK$ is parallel to the axis, and $AK$=$KB$, then $BK$ is an ordinate to diameter $MK$ and $$\frac{MK}{ML}=\frac{BK^2}{FL^2}$$ This is the fundamental proportion of the parabola (see Apollonius, Conics, I, 20). Thus by Archimedes' $Prop. 2$, $MK$=$MP$. Therefore, $M$ lies on $JB$, so that by similar triangles $$\frac{MK}{ML}=\frac{BM}{MN}$$ and $$\frac{BK}{FL}=\frac{BK}{EK}=\frac{BM}{MG}$$ Therefore by substitution$$\frac{BM}{NM}=\frac{BM^2}{MG^2}$$  Hence $BM, MG, MN$ are in continued proportion:$$\frac{BM}{MG}=\frac{MG}{MN}=\frac{BM+MG}{MG+MN}=\frac{BG}{NG}$$ But by similar triangles$$\frac{BG}{NG}=\frac{EG}{FG}$$and$$\frac{BM}{MG}=\frac{BK}{KE}$$Therefore, $$\frac{BK}{KE}=\frac{EG}{FG}$$$QED$ for $Prop. 4$.     From this, and the fact that $AD=DC$ and $EG=GH$, Archimedes proves $Prop. 5$: $$\frac{BE}{AE}=\frac{HF}{EF}$$Since from $Prop. 4$$$\frac{BK}{BK-KE}=\frac{EG}{EG-FG}$$then$$\frac{BK}{AE}=\frac{EG}{EF}$$whence $$\frac{2BK}{AE}=\frac{2EG}{EF}$$so that$$\frac{AB}{AE}=\frac{EH}{EF}$$hence$$\frac{AB-AE}{AE}=\frac{EH-EF}{EF}$$or$$\frac{BE}{AE}=\frac{HF}{EF}$$which proves $Prop. 5$. But by similarity$$\frac{BE}{AE}=\frac{BG}{DG}$$Therefore$$\frac{HF}{EF}=\frac{BG}{DG}$$and so$$\frac{HF+EF}{EF}=\frac{BG+DG}{DG}$$or$$\frac{EH}{EF}=\frac{DB}{DG}$$Alternating this gives the needed proportion$$\frac{EF}{DG}=\frac{EH}{JD}$$since $JD=DB$.
A: 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 $JD=BD$.
