Area of a trapezoid with perpendicular diagonals, embedded in a triangle Let $ABED$ be a trapezoid as in

If $AB \parallel DE$, $AE \perp BD$, $AB = 10$, $DE = 4$ and $\angle ACB = 45°$, what's the area of $ABED$?
I must also mention that this is for an elementary geometry course, so I have restricted myself to that approach --I have not attempted anything related to calculus or analytic geometry for the same reason.
I have tried the following: (1) Using triangle similarity to get the height of the trapezoid (2) Using sine or cosine law to the same end (3) Trying to get the length of its diagonals with similar approaches as above (4) Using some weird formula for the area of a triangle by multiplying its three sides and dividing it by the circunradius. None of the above worked. Every time I get equations with more than one variable, and can't figure out a way to get a system that would help me solve them.
Thank you all for any help!
 A: The hint.
Let $FE=x$, $DF=y$, $L\in AB$ such that $DL||EB.$
Thus, $$x^2+y^2=16,$$
$$AF=2.5x,$$ $$BF=2.5y$$ and since $DL=EB$, $AL=10-4=6$ and $\measuredangle ADL=45^{\circ},$ by low of cosines we obtain:
$$AD^2+LD^2-2AD\cdot LD\cos45^{\circ}=AL^2$$ or
$$6.25x^2+y^2+6.25y^2+x^2-2\sqrt{(6.25x^2+y^2)(6.25y^2+x^2)}\cdot\frac{1}{\sqrt2}=36.$$
After solving of this system use $$S_{ABED}=\frac{1}{2}AE\cdot DB.$$
I got $xy=\frac{160}{21}$ and $S_{ABED}=\frac{140}{3}.$
A: Ok so using the fact, that many triangles are congruent, I constructed this:

Now:
$x^2 + y^2 = 4$ (1)
$4z^2 + 4m^2 - 4*\sqrt{2}mz = 16 => mz = \frac{z^2+m^4 -4}{\sqrt{2}}$ (2)
$4y^2 +25x^2 = 9m^2$ (3)
$4x^2 + 25y^2 = 9z^2$ (4)

Add up (3) and (4), then replace $z^2 + m^2$ in (2) and calculate $mz = \frac{40*\sqrt{2}}{9}$
Now, using the area formula, which involves sinus: $S = S_{big} - S_{small} = 0.5*25*mz*\sin{45} - 0.5*4*mz*\sin{45} = \frac{0.5*21*40*\sqrt{2}}{\sqrt{2}*9} = \frac{140}{3}$
A: If $\measuredangle FAB=\theta^{\circ}$ then $AE = 14 \cos \theta$, $DB = 14 \sin \theta$, and the area we require is $98 \sin \theta \cos \theta$. I can furthermore construct such a diagram for any $\theta$ (construct $AB$ then $F$ then $E$, $D$ and $C$). 
The problem is underspecified.
eta: Apologies, I was not seeing the other part of the specification. The angle at $C$ is given. That makes the problem solvable.
