Showing that the area of a parabolic sector is half the area of a corresponding region bounded by the directrix (without Calculus) 
Given parabola:

It is necessary to prove that the area of the parabolic sector (green) is equal to half the area of the parabolic rectangle (orange) without calculus. ("Something like" Archimedes' Quadrature of the Parabola may be used, although (according to a comment made to an answer) not the Quadrature itself.)

I know I need to use the optical property of a parabola, but how exactly?
Any thoughts would be so much appreciated.
Original text of the problem from the book "Measurement" (page 190) by Paul Lockhart:
"Using the method of exhaustion, Archimedes was able to show that the area of the sector is exactly half that of the rectangle. Can you do the same? Why is the area of a parabolic sector equal to half the area of the parabolic rectangle?"
 A: Let $F$ be the focus of the ellipse, $A$ and $B$ the endpoints of the chord, $H$ and $K$ their projections on the directrix, $C'$ the midpoint of $AB$, $C$ the intersection of the parabola with the line through $C'$ perpendicular to the directrix (see diagram).
Upper area $A_u$ is the sum of the area of parabolic segment $ABC$ and the area $A_{ABF}$ of triangle $ABF$ (if the parabolic segment contains $F$ then $A_{ABF}$ must be subtracted, but for simplicity I won't consider this case in the following). Lower area $A_l$ is the difference between the area $A_{ABKH}$ of trapezoid $ABKH$ and the area of parabolic segment $ABC$. In turn, the area of parabolic segment $ABC$ is $4/3$ the area $A_{ABC}$ of triangle $ABC$ (by Archimedes' theorem).
We must show that $A_l=2A_u$, that is: 
$$
A_{ABKH}-{4\over3}A_{ABC} =2\left( A_{ABF}+{4\over3}A_{ABC}\right),
$$
or:
$$
A_{ABKH}=2A_{ABF}+{4}A_{ABC}.
$$
A direct (but ugly) way to check this equality is that of considering a parabola $y=x^2$ and express all areas in terms of the abscissae of $A$ and $B$. I can give details if needed.

A: I'll try to give a proof which should be on the right track.
Let $P$ and $Q$ be two points on the parabola, $M$ and $N$ their projections on the directrix, $F$ the focus. When $Q$ approaches $P$ the area of trapezoid $PQNM$ becomes the double of the area of triangle $PQF$, because bases $PM$ and $PF$ are equal, $QN$ approaches $PM$, and the ratio of altitudes $QH$ and $QK$ tends to $1$, because line $PQ$ becomes by definition the tangent at $P$, which is the bisector of angle $\angle FPM$ (focal property of parabola).
The area of the parabolic sector is the sum of triangles like $PQF$, while the area of the parabolic rectangle is the sum of trapezoids like $PQNM$, hence the thesis follows.

A: Let the parabola have focus $F$, and let the sector be determined by the arc between points $A$ and $B$. Let $F^\prime$, $A^\prime$, $B^\prime$ be the feet of the perpendiculars from these points to the directrix. Finally, let $C$ and $C^\prime$ be a point on the parabola, and its directrix projection, such that $C^\prime$ is the midpoint of $\overline{A^\prime B^\prime}$.

We will use Archimedes' Quadrature of the Parabola (because the restriction against it was not stated until after I'd done this work):
$$|\,\text{parabolic sector } ACB\,| = \frac43\,|\,\triangle ABC\,| \tag{0}$$
so that our task is to show
$$2\left(|\triangle FAB| + \frac43|\triangle ABC| \right) = |\square ABB^\prime A^\prime| - \frac43|\triangle ABC| \tag{1}$$
Since
$$\begin{align}
|\triangle FAB| &= |\square FBB^\prime F^\prime| - |\square FAA^\prime F^\prime| - |\square ABB^\prime A^\prime| \\
|\triangle ABC| &= |\square ABB^\prime A^\prime| - |\square ACC^\prime A^\prime| - |\square CBB^\prime C^\prime|
\end{align} \tag{2}$$
we see that condition $(1)$ becomes
$$|\square ABB^\prime A^\prime| - 4 |\square ACC^\prime A^\prime| - 4 |\square BCC^\prime B^\prime| - 2 |\square FAA^\prime F^\prime| + 2 |\square FBB^\prime F^\prime| = 0 \tag{3}$$
Define
$$a := |F^\prime A^\prime| \qquad b := |F^\prime B^\prime|$$
$$2f := |FF^\prime| \qquad h = |AA^\prime| \qquad j := |BB^\prime| \qquad k := |CC^\prime|$$
with $a$ and $b$ being signed distances (in anticipation of coordinatizing later), and $b\geq a$. With these, the areas of various trapezoids are readily determined, so that the left-hand side of $(3)$ becomes
$$\begin{align}
&\phantom{=\;\;}\frac12(b-a)(h+j) - (b-a)(h+k)-(b-a)(j+k) -a(2f+h) + b(2f+j) \\
&= 2 (b - a) (f - k) + \frac12 (a + b) (j - h) \tag{4}
\end{align}$$
Coordinatizing, with the $x$- and $y$-axes as the parabola's directrix and axis, the parabola's equation is $y = \dfrac1{4f}x^2 + f$, and we have
$$h = \frac1{4f}a^2 + f \qquad j = \frac1{4f} b^2 + f \qquad k = \frac1{4f}\left(\frac{a+b}{2}\right)^2 + f = \frac1{16f}(a+b)^2+f \tag{5}$$
whence
$$\begin{align}
j-h &= \frac1{4f}\left(b^2-a^2\right) = \frac1{4f}(b-a)(b+a) \\[4pt]
k-f &= \frac1{16f} (a+b)^2
\end{align}$$
and we see that $(4)$ vanishes, verifying $(3)$ (and thus $(1)$). $\square$
