Area calculation of triangels in a parallelogram My problem is as follow:
Let ABCD be a parllelogram, with points E and F on sides AD and BC so that the length of BF and DE are the same. Moreover, a point K is arbitrarily placed on side CD. Three lines are drawn in the parallelogram: line AK, line BK and line EF. The point where line AK and EF intersects are called U. The point where line BK and EF intersect are called W. 
Now to the question: How can I prove that the area of (AEU) + the area of (BFW) = the area of triangel (UWK)?
At this moment I have no idea where to start, so some insightful thoughts would be appreciated. 
/Alex
 A: Hint: Take a point $Y$ on $AB$ such that $KY||AD$. Let $KY$ intersect $EF$ at X. Show:
$$\Delta AEU\sim\Delta KUX; \Delta BFW\sim \Delta KWX.$$
Addendum: If the points $E$ and $F$ are midpoints, then the areas are indeed equal:
$\hspace{4cm}$
Otherwise the claim is not true. WLOG, consider the figure:
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Using similarities of triangles:
$$\begin{cases} \frac{t}{1}=\frac{z+4}{z} \\ \frac 41=\frac{z+10}{z} \end{cases} \Rightarrow \begin{cases} t=\frac{11}{5} \\ z=\frac{10}{3} \end{cases}.$$
Again using the similarities of triangles:
$$\begin{cases} \frac{S_1}{S_2}=\left(\frac{4}{\frac{11}{5}}\right)^2 \\ \frac{S_4}{S_3}=\left(\frac{1}{\frac{11}{5}}\right)^2 \end{cases} \iff \begin{cases} S_1=\frac{400}{121}S_2 \\ S_4=\frac{25}{121}S_3 \end{cases} \iff S_1+S_4\ne S_2+S_3.$$
Note: The relationship between the areas can be established in terms of the positions of the points $E$ and $K$.
A: I have a proof which requires some calculation. Nevertheless it solves the problem.

Suppose AE=CF=t, and AD=BC=b, KG=x (where KG//AD//BC, and G is the intersection of KG and EF), then we have DE=BF=b-t.
We want to prove that $S_{\triangle AEU}+S_{\triangle BFW}=S_{\triangle UWK}$, which is equivalent to prove that
$$
\frac 1 2 AE \cdot EU \cdot \sin\angle AEU + \frac 1 2 BF \cdot FW \cdot \sin\angle BFW = \frac 1 2 UW \cdot GK \cdot \sin\angle UGK
$$
Since the lines are parallel, the sine of the three angles are equal. Thus what we are looking for is that
$$
AE \cdot EU + BF \cdot FW = UW \cdot GK
$$
Assume that $AB\nparallel EF$. Using the property of parallel lines, it is easy to find out that 
$$
\begin{align}
EG:GF = (KG-DE):(CF-KG) &= (b-x-t):(x-t) \\
EU:UG &= t:x \\
GW:WF &= x:(b-t)
\end{align}
$$
Now we shall integrate these ratios. Multiply the first equation by $(x+t)(x+b-t)$, we have
$$
\begin{align}
EG:GF       &= (b-x-t)(x+t)(b+x-t) : (x-t)(x+t)(b+x-t) \\
EU:UG:GW:WF &= t(b-x-t)(b+x-t) : x(b-x-t)(b+x-t) \\
            &: x(x-t)(x+t) : (b-t)(x-t)(x+t)
\end{align}
$$
Suppose $EU = \gamma \cdot t(b-x-t)(b+x-t)$, then
$$
\begin{align}
AE \cdot EU + BF \cdot FW &= t\cdot \gamma \cdot t(b-x-t)(b+x-t) + (b-t) \cdot \gamma \cdot x(b-x-t)(b+x-t) \\
&= \gamma\left( t^2 (b-t)^2 - x^2 t^2 + x^2 (b-t)^2 - t^2 (b-t)^2 \right) \\
&= \gamma\left(x^2 (b-t)^2 - x^2 t^2 \right) \\\\
UW\cdot GK &= x\cdot \gamma \cdot \left(x(b-x-t)(b+x-t)+x(x-t)(x+t) \right) \\
&= \gamma\left(x^2 (b-t)^2 - x^2 t^2 \right)
\end{align}
$$
which follows $AE \cdot EU + BF \cdot FW = UW \cdot GK$ and finishes the proof.
The circumstance where EF//AB may be discussed separately, which is indeed easier than the general case.
