Dividing Land which is an Irregular quadrilateral A piece of land was left to two sisters,Ella and Bella,by their parents in their will.The land was shaped as an irregular quadrilateral (ABCD is and irregular quadrilateral with E as a point where the two diagonals of the quadrilateral meets).The will declared that the land should be divided equally among the two daughters.Ella suggested dividing the land into four regions by constructing the two diagonals.Ella would take two regions (AED and BEC),leaving Bella with the regions ABE and CED.Bella thought that seemed fair until she talked to her lawyer and discovered that the sums of the regions were not equal.When she pressed Ella ,Ella replied,"That's true ,but the products of the region are equal!"Bella did not know what to do.
Are the products of the region equal?Model this problem on sketchpad and find out whether Ella's claim is true.If it is true ,prove it ,if not ,explain why not?
 A: The quadrilateral must be convex, else the proposed division would make no sense. Suppose that the vertices $A$, $B$, $C$, $D$ are given in counterclockwise order. We first sketch the quadrilateral.  
Consider the diagonal $AC$.  We first look at triangles $AEB$ and $CEB$. If we think of them as having bases $AE$ and $CE$, then they have the same height. It follows that their areas are in the same ratio as their bases,  and therefore 
$$\frac{|AEB|}{|CEB|}=\frac{|AE|}{|CE|}.$$
(We are using the abbreviations $|XYZ|$ for the area of triangle $XYZ$, and $|XY|$ for the length of line segment $XY$.)
Similarly, by considering triangles $AED$ and $CED$, again with bases $AE$ and $CE$, we obtain
$$\frac{|AED|}{|CED|}=\frac{|AE|}{|CE|}.$$
Thus
$$\frac{|AEB|}{|CEB|}=\frac{|AED|}{|CED|}.$$
It follows that $|AEB||CED|=|CEB||AED|$.
So the products are indeed equal.  
A: Let $\theta$ be the angle $\leq\pi/2$ that the two diagonals make with one another.  Then the areas of the four triangles are $\frac{1}{2}\overline{AE}\thinspace\overline{BE} \sin\theta$, $\frac{1}{2}\overline{BE}\thinspace\overline{CE} \sin\theta$, $\frac{1}{2}\overline{CE}\thinspace\overline{DE} \sin\theta$ and $\frac{1}{2}\overline{DE}\thinspace\overline{AE} \sin\theta$.  So both products are $\frac{1}{4}\overline{AE}\thinspace\overline{BE}\thinspace\overline{CE}\thinspace\overline{DE}\sin^2\theta$.
A: It is true. Here is an ugly but simple coordinate-based proof. Take the point $D=(0,0)$. An invertible linear transformation will leave the ratios of areas unchanged, so apply a transformation so that results in $A=(0,1)$, $C=(1,0)$. Then only $B=(x,y)$ is arbitrary.
The point $E$ is at the intersection of $AC$ and $DB$, so can be easily computed to be $E=(\frac{x}{x+y}, \frac{y}{x+y})$.
We have the area formulae:
$$2\;\mathbb{area}(AED) =
|\det \left( \begin{array}{cc}
0 & 1 \\
\frac{x}{x+y} & \frac{y}{x+y}
\end{array} \right) | = \frac{x}{x+y}$$
$$2\; \mathbb{area}(ABE) =
|\det \left( \begin{array}{cc}
0 & 1 \\
x & y
\end{array} \right) | -2\;\mathbb{area}(AED)= \frac{x^2+xy -x}{x+y}$$
$$2\;\mathbb{area}(CED) =
|\det \left( \begin{array}{cc}
1 & 0 \\
\frac{x}{x+y} & \frac{y}{x+y}
\end{array} \right) | = \frac{y}{x+y}$$
$$2\; \mathbb{area}(BEC) =
|\det \left( \begin{array}{cc}
1 & 0 \\
x & y
\end{array} \right) | -2\;\mathbb{area}(CED)= \frac{y^2+xy-y}{x+y}$$
A quick multiplication shows the assertion to be correct.
