I have tried solve the following problem, but I can not see how to approach it:
In the figure, $ABCD$ is an arbitrary convex quadrilateral, $M$ and $N$ are, respectively, the middle points of $\overline{AB}$ and $\overline{CD}$, and $S,S_1,S_2$ are the area of the shaded regions. Prove that $$S=S_1+S_2$$
Any hints are welcome!

I could not draw the diagram but here is the solution. Draw a dotted line from A parallel to DC. Draw dotted perpendiculars from A to DC, M to DC and also from B to DC, they intersect DC at K, G and H respectively (say). Now MG and BH also intersect the line drawn from A parallel to DC say at point E and F respectively. Now as AM = MB so AE = EF and ME = (1/2) BF, let ME = x, BF = 2x. Now say AN and MD intersect at S1 and MC and BN intersect at S2. Now
$\Delta ADN = (1/2){\overline{DN}}.{\overline{AK}}$,
$\Delta MDN = (1/2){\overline{DN}}.{\overline{MG}} = (1/2){\overline{DN}}.({\overline{AK}} + x) $,
$\Delta MNC = (1/2){\overline{NC}}.({\overline{AK}} + x) = (1/2){\overline{DN}}.({\overline{AK}} + x)$,
$\Delta BNC = (1/2){\overline{DN}}.({\overline{AK}} + 2x)$,
Now $\Delta ADN = \Delta ADS_1+ \Delta S_1DN$,
$\Delta MDN = \Delta S_1DN + \Delta MS_1N$,
$\Delta MNC = \Delta MS_2N + \Delta S_2NC$,
$\Delta BNC = \Delta S_2NC + \Delta BS_2C$
So, $\Delta MDN - \Delta ADN = \Delta MS_1N - \Delta ADS_1 = (1/2){\overline{DN}}x$
$\Delta BNC - \Delta MNC = \Delta BS_2C - \Delta MS_2N = (1/2){\overline{DN}}x$
So $\Delta MS_1N + \Delta MS_2N = \Delta BS_2C + \Delta ADS_1$
Note: I assumed BH > AK , but anyway it does not matter.