These are two questions that have come up to my mind. I do not want to make 2 different questions, because the second one is quite short and might be very simple to answer.
1) $ABCD$ is a quadrilateral and $E$ is the intersection between diagonals $\overline{AC}$ and $\overline{BD}$. Assume that $\triangle BEC$ and $\triangle AED$ have the same area. Then do $\triangle ABE$ and $\triangle CED$ have the same area?
What I did was to use the formula for the area of a triangle with sides $\overline{a}$, $\overline{b}$, and $\overline{c}$, and $\angle A$ opposite to $\overline{a}$:
$$\text{Area} = \frac{bc \sin (A)}{2}$$
With that I can't get that $BE \cdot EC = AE \cdot ED$. Then, with that you can show that $\overline{AB} || \overline{CD}$ and then you can show that $[\triangle ABE] = [\triangle CED]$. Is there any simpler way to get this conclusion?
Claim 1) is not true. Consider trapezoid.
$A_{\triangle{ABC}}=A_{\triangle{BCD}}$ (triangles with the same base between parallel lines).
Hence, $A_{\triangle{ABE}}=A_{\triangle{CDE}}$ but $A_{\triangle{BCE}} \ne A_{\triangle{AED}}$