I´ve some problem with the area calculation in a convex quadrilateral. My problem looks like this:
Let $ABCD$ be a convex quadrilateral with points $X$ and $Y$ on side $AB$ so that $AX = XY = YB$, and with points $Z$ and $W$ on side $CD$ so that $CZ = ZW = WD$. Now I want to prove that the area of the field $XYZW$ is equal to one third $(1/3)$ of the area of the convex quadrilateral $ABCD$.
I don´t know how to solve this problem. Does anyone have an idea?
/Alf
Here is a proof of the assertion. I'll start with a simpler statement:
Lemma: Let $P_1$, $P_2$, $P_3$ be three points in a straight line, such that $|P_1P_2|=|P_2P_3|$, and let $l$ be some other line that does not intersect $P_1P_3$.
Drop perpendiculars from each of the three points onto the line $l$, and name their lengths $l_1$, $l_2$, and $l_3$.
Then we have $l_2 = \frac{l_1+l_2+l_3}{3}$.
Proof:
It is fairly obvious from the picture that $l_2-l_1 = l_3-l_2$. This is because $P_1P_2$ has the same length and slope as $P_2P_3$, so the line through the points rises by the same amount over the two segments, and hence $l_2-l_1 = l_3-l_2$. To prove it without using a coordinate system, you can simply show that the two right-angled triangles are congruent.
From $l_2-l_1 = l_3-l_2$ it easily follows that $l_2 = \frac{l_1+l_2+l_3}{3}$.
Note that this equation still holds if the line slopes downwards, since the proof works just as well in mirror image. Not only that, you can check that the equation even holds if the line is horizontal or vertical.
We can use the above lemma to show that the areas of the three triangles in the following picture satisfy $T_2 = \frac{T_1+T_2+T_3}{3}$
The reason is simple. From the lemma, the heights of the three triangles satisfy $l_2 = \frac{l_1+l_2+l_3}{3}$. They have the same base length, so by multiplying this equation by half the base length you get the same equation but for their areas: $T_2 = \frac{T_1+T_2+T_3}{3}$.
Now we can look at what happens with the quadrangles in the original question. Here's a picture:
Add the same diagonal to each of the the three quadrangles. This splits the problem into two sets of three triangles. One set satisfies $T_2 = \frac{T_1+T_2+T_3}{3}$, but by turning the picture upside down you can see that the other set similarly satisfies $S_2 = \frac{S_1+S_2+S_3}{3}$.
By adding these together it then follows that $S_2+T_2 = \frac{S_1+T_1+S_2+T_2+S_3+T_3}{3}$, or $XYZW = \frac{ABCD}{3}$, as required.