Relating the areas of regions determined by joining midpoints of two sides of a convex quadrilateral to vertices

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!

• Do you mean "an arbitrary convex quadrilateral"? – ajotatxe Jun 12 '18 at 18:51
• Yes, It should be a convex quadrilateral. I will change the statement. – DiegoMath Jun 12 '18 at 18:57

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.

Using Shoe Lace formulas for quadrilateral and triangle, you can verify algebrically the sum of areas of triangle is area of the quadrilateral.

A = [x1,y1]; B = [x2,y2]; C = [x3,y3]; D = [x4,y4]

M = (1/2*x1 + 1/2*x2, 1/2*y1 + 1/2*y2)

N = (1/2*x3 + 1/2*x4, 1/2*y3 + 1/2*y4)

Q = [(x1*x3*y1 + x2*x3*y1 - x1*x4*y1 + x2*x4*y1 - x3*x4*y1 - x4^2*y1 - 2*x1*x4*y2 + x3*x4*y2 + x4^2*y2 - x1^2*y3 - x1*x2*y3 + 2*x1*x4*y3 + x1^2*y4 + x1*x2*y4 - x1*x3*y4 - x2*x3*y4 + x1*x4*y4 - x2*x4*y4)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4), (x3*y1^2 - x4*y1^2 + x3*y1*y2 - x4*y1*y2 - x1*y1*y3 + x4*y1*y3 - x1*y2*y3 + x4*y2*y3 + x1*y1*y4 + 2*x2*y1*y4 - 2*x3*y1*y4 - x4*y1*y4 - x1*y2*y4 + x4*y2*y4 + x1*y3*y4 - x2*y3*y4 + x1*y4^2 - x2*y4^2)/(2*x2*y1 + x3*y1 - 3*x4*y1 - 2*x1*y2 + x3*y2 + x4*y2 - x1*y3 - x2*y3 + 2*x4*y3 + 3*x1*y4 - x2*y4 - 2*x3*y4)]

R = [(2*x2*x3*y1 - x3^2*y1 - x3*x4*y1 - x1*x3*y2 + x2*x3*y2 + x3^2*y2 - x1*x4*y2 - x2*x4*y2 + x3*x4*y2 - x1*x2*y3 - x2^2*y3 + x1*x3*y3 - x2*x3*y3 + x1*x4*y3 + x2*x4*y3 + x1*x2*y4 + x2^2*y4 - 2*x2*x3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4), (x3*y1*y2 - x4*y1*y2 + x3*y2^2 - x4*y2^2 + x2*y1*y3 - x3*y1*y3 - 2*x1*y2*y3 - x2*y2*y3 + x3*y2*y3 + 2*x4*y2*y3 + x1*y3^2 - x2*y3^2 + x2*y1*y4 - x3*y1*y4 + x2*y2*y4 - x3*y2*y4 + x1*y3*y4 - x2*y3*y4)/(2*x2*y1 - x3*y1 - x4*y1 - 2*x1*y2 + 3*x3*y2 - x4*y2 + x1*y3 - 3*x2*y3 + 2*x4*y3 + x1*y4 + x2*y4 - 2*x3*y4)]