Areas of triangles in hexagon 
A hexagon $ABCDEF$ with parallel opposite sides is given. Prove that $[ACE]=[BDF].$
  (here $[]$ denotes area of triangle)


Since the sides are parallel does that mean that it is equiangular as well? If so how do we prove that and how does that help us solve the problem? I am not seeing an easy way to show the areas are equal.
 A: If three points $\mathrm{A}_1(a_1,\ b_1),\ \mathrm{A}_2(a_2,\ b_2),\ \mathrm{A}_3(a_3,\ b_3)$ are known, the area can be calculated.
$$\triangle\mathrm{A_1A_2A_3}= \frac{1}{2}|x_1y_2+x_2y_3+x_3y_1-x_2y_1-x_3y_2-x_1y_3|$$
The formula can be proved by the fact $\operatorname{distance}(ax+by+c=0,\ (x_0, y_0)) = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$.
Let's simplify the situation.
$$\overline{\mathrm{DE}} : y = 0,\ \overline{\mathrm{AB}} : y = 1,\ \overline{\mathrm{EF}} : y = ax,\ \overline{\mathrm{BC}} : y = ax+b,\ \overline{\mathrm{CD}} : y = cx+d,\ \overline{\mathrm{FA}} : y = cx+e.$$
Then the following is true.
$$\mathrm{D}(-\frac{d}{c},\ 0),\ \mathrm{E}(0,\ 0),\ \mathrm{F}(\frac{e}{a-c},\ \frac{ae}{a-c}),\ \mathrm{A}(\frac{1-e}{c},\ 1),\ \mathrm{B}(\frac{1-b}{a},\ 1),\ \mathrm{C}(\frac{d-b}{a-c},\ \frac{ad-bc}{a-c}).$$
Therefore
$$\triangle \mathrm{ACE} = \frac{1}{2}|\frac{1-e}{c}\frac{ad-bc}{a-c}-\frac{d-b}{a-c}|=\frac{1}{2}|\frac{-ade+ad+bce-cd}{ac-c^2}|
\\ \triangle \mathrm{BDF} = \frac{1}{2}|-\frac{d}{c}\frac{ae}{a-c}+\frac{e}{a-c}+\frac{d}{c}-\frac{1-b}{a}\frac{ae}{a-c}|=\frac{1}{2}|\frac{-ade+ad+bce-cd}{ac-c^2}|$$
and proved.
The picture shows reducing gap or rotating the parallel lines in the case of regular hexagon. Perhaps, you may solve the problem without coordinate system considering properties such as below.
$\hskip2in$
A: A suitable affine transformation $T$ of your figure will make two of the parallels vertical and two of the parallels horizontal. This $T$ will multiply all areas with the same factor. Therefore you may assume that two of the parallels are $x=0$, $x=1$, and that two of the parallels are $y=0$ and $y=1$. It follows that two of the vertices are $A=(1,0)$ and $D=(0,1)$, and two more are  $B=(1,b)$ and $C=(c,1)$ with  $0<c<1$ and $0<b<1$. Since $E$ and $F$ have to lie on the $y$- and the $x$-axis respectively, and $E\vee F$ has to be parallel to $B\vee C$ it follows that
$$E=\lambda(0,1-b)), \qquad F=\lambda(1-c,0)$$
for a certain $\lambda>0$. Now compute
$$2|\triangle(ACE)|=(C-A)\wedge(E-A)=(c-1,1)\wedge\bigl(-1,\lambda(1-b)\bigr)$$
and
$$2|\triangle(DFB)|=(F-D)\wedge(B-D)=\bigl(\lambda(1-c),-1\bigr)\wedge(1,b-1)\ ,$$
and verify that the values are the same.
