# A property of the area of quadrilaterals

I did not know this property of quadrilaterals proposed me a student. I could not prove to the first attempt but I could do after devoting some time. I want to share it here to see if anyone has any proof different from mine.

In the four vertices of a quadrilateral $ABCD$ whose area is $S$, they are drawn parallel to two rectangular straight lines which determines two rectangles (green in the figure below) whose areas are $R$ and $r$. Prove that $$R + r = 2S$$

Explanation of parallel lines: The quadrilateral and a pair of orthogonal directions are given. $A$ and $C$ are opposite vertices. For one rectangle, draw "horizontal" lines through $A$ and $C$ and "vertical" lines through $B$ and $D$. For the other rectangle, draw "vertical" lines through $A$ and $C$ and "horizontal" lines through $B$ and $D$.

• You should show your proof so that we don't duplicate your effort. – Blue Aug 27 '16 at 22:35
• I have the proof but had intended to publish later. I will publish tomorrow because if the posting at this time, I will not have the different possible answers I would like to know. – Piquito Aug 27 '16 at 22:43
• How do we know our answers are different than yours if we don't know what yours is? If someone tells you something you already know, then you've wasted their (and your) time; that's not very considerate. I guess I'll wait until you publish your solution tomorrow before I start thinking about this problem. – Blue Aug 27 '16 at 22:48
• The way "they are drawn parallel to two rectangular straight lines which determines two rectangles " looks to me very very fuzzy. Could you explain a little more how this rectangles are built ? – Jean Marie Aug 27 '16 at 23:08
• @Blue - with the OP's disclaimer, I believe that concern is lessened. Only people who like the challenge and don't mind finding the OP's solution should even consider solving it; for the others, there are hundreds of other problems on this site they can help posters with. – mathguy Aug 27 '16 at 23:31

There is an immediate way to find the area of any plane polygon and this gives for our quadrilateral $ABCD$ $$2S =(x_D-x_A)(y_A+y_D)+(x_C-x_D)(y_C+y_D)-(x_B-x_A)(y_B+y_A)-(x_C-x_B)(y_C+y_B)$$ which simplifies to$$2S=x_C\cdot y_D-x_C\cdot y_B-x_A\cdot y_D+x_A\cdot y_B+x_B\cdot y_C-x_B\cdot y_A-x_D\cdot y_C+x_D\cdot y_A$$ But this is just $$(y_D-y_B)(x_C-x_A)+(y_C-y_A)((x_B-x_D)$$ which is the apparent area of the two rectangles.
• +1. It's important to note that, with such symbolic generality, you're playing with signed areas, which requires care in applying the result. For example, take $A (0,0)$, $B(2,0)$, $C(-1,2)$, $D(-3,2)$. Quad $ABCD$ is a parallelogram w/base & height $2$; the "green" rectangles have dimensions $5\times 2$ and $1\times 2$. But the quad's double-area, $8$, is clearly not the sum of $10$ & $2$ ---as one might expect from $2S = R + r$--- rather, it's the difference. The trick is that formula takes the smaller rect's area to be negative two, so that the sum $2\cdot 4 = 10+(-2)$ works. – Blue Aug 28 '16 at 2:18
As you move $C$ along the straight line that passes through $C$ and is parallel to $BD$, the area of the quadrilateral does not change. The area of one rectangle increases and the area of the other decreases. To see that the increase and the decrease offset each other exactly, express the areas explicitly in terms of coordinates; then, to prove the fact that the offset is exactly zero, the condition is exactly that the point $C$ moves parallel to $BD$ (the slope of $BD$ is the same as the slope of $CC'$ where $C'$ is the new position of $C$).
So in order to prove the result in general, we may move $C$ parallel to $BD$ until we find a position where we can prove the result. In particular, it is enough to consider the case when $C$ is on the extension of $AD$ beyond $D$.
Then, with $C$ so moved, we can apply the argument equally to $D$: we can move it parallel to $AC$ (except in this case $D$ is on $AC$ already - so $D$ is actually moving along $AC$). In particular, it suffices to prove the result when $D$ is on top of $A$. Finally, move $B$ parallel to $AC$ until it is on the same "horizontal" line as $A$ (which is the same as $D$ now). In that case one rectangle has area zero, and the result is trivial. Since moving a vertex parallel to the diagonal it is not on does not change the truth of the statement, the statement is proved in full generality.