I have solved this problem by use of the Cartesian plane, but the solution is long and I am sure that I have overkilled it and that there is a simpler solution...
Based on where this question came from, I believe it is possible to solve it using basic knowledge area formulae.
[Edit: Construct the lines AC, BF CE and BD. The lines BF and AE intersect at G and the lines CE and BD intersect at H.]
Triangles $\triangle BCH$ and $\triangle DEH$ are similar. The lengths $|BC| = \frac12$ and $|DE| = \frac34,$ so $|CH|:|EH| = |BC|:|DE| = 2:3$ and $|CH|:|CE| = 2:5.$ (Here $x:y$ is just a way of saying "the proportion of $x$ to $y$"; rewrite it $x/y$ if the notation is too old-fashioned for you.)
As a result, the perpendicular distance from $H$ to the nearest point on $BC$ is $\frac25$ and the area of $\triangle BCH$ is $\frac12 bh = \frac12\left(\frac12\right)\left(\frac25\right) = \frac1{10}.$
Triangles $\triangle ABG$ and $\triangle EFG$ also are similar, but are in the proportion $2:1,$ so the perpendicular distance from $G$ to $AB$ is $\frac23$ and the area of $\triangle ABG$ is $\frac12\left(\frac12\right)\left(\frac23\right) = \frac16.$
The triangle $\triangle ACE$ (with area $\frac12$) can be partitioned into three disjoint pieces: the triangle $\triangle ABG,$ the triangle $\triangle BCH,$ and the quadrilateral $BHEG.$ To get the area of the third piece (the quadrilateral), subtract the known areas of the two smaller triangles from the area of $\triangle ACE.$
Note that you can fairly easily find the distances of $G$ and $H$ from the sides $AF$ and $CD,$ which would be tantamount to finding complete Cartesian coordinates for every labeled point in the figure (given the information above), but you do not need those coordinates.
(This is quite different from my earlier answer because important details of the problem were later provided.)
