How can this euclidean geometry problem be solved? Two straight lines parallel to the base of a trapezoid divide each lateral side into three equal parts. The entire trapezoid is separated by the lines into three parts. Find the area of the middle part if the areas of the upper and lower parts are $S1$ and $S2$, respectively.
The answer is $\dfrac12 (S1+S2)$. I've tried to attack this problem in countless possible ways, but failed to prove it. Help will be much appreciated.
 A: Given a trapezoid $ABCD,$ suppose we construct a congruent trapezoid $CDEF$ with bases collinear with the bases of the original trapezoid,
and then we subdivide both trapezoids by placing straight lines parallel
to the bases, dividing each side into three lateral parts, as shown here:

By symmetry, the small trapezoids colored blue are congruent
with area $S_1$ each, the small trapezoids colored yellow are congruent with area to be determined, and the small trapezoids colored orange are congruent with area $S_2$ each.
But the trapezoids $ABCD$ and $CDEF$ together form a parallelogram
$ABFE,$ which is divided into three congruent parallelograms
by the same lines that subdivide the trapezoids.
The upper parallelogram (colored blue and orange) is equal in area to
the sum of the upper and lower parts of trapezoid $ABCD$
(colored blue and orange, respectively)
and also equal in area to the yellow parallelogram,
which has twice the area of the middle part of trapezoid $ABCD$.
Hence the middle (yellow) part of trapezoid $ABCD$ has an area
half the sum of the other two parts (blue and orange),
namely $\frac12(S_1 + S_2).$
You might recognize this as a simple adaptation of the proof of the formula for the area of a trapezoid.
A: Short answer:
The sides of the three parts are in arithmetic progression, and so are the areas (proportional to the averages of successive sides).

Longer answer:
Let the parallel sides have length $a$ and $a+3d$, and let the height be $6h$.
Then by Thales, the sides of the three parts are $a,a+d$, $a+d,a+2d$ and $a+2d,a+3d$, and the heights are $2h$.
$$S_1=h(a+a+d),S_2=h(a+d+a+2d),S_3=h(a+2d+a+3d).$$
