Parallelogram and Areas Consider the parallelogram $ABCD$. On sides $BC$ and $CD$ take points $E$ and $F $ respectively such that $\frac{BE}{EC} = \frac{CF}{ FD}$. If the segments $AE$ and $AF$ cut $BD$ at $K$ and $L$, show that $(AKL)=(BEK)+(DLF)$


Notice first that $(BEK)=(ABE)-(ABK)$, $(DLF)=(ADF)-(ADL)$,
$(AKL)=(ABD)-(ABK)-(ADL)$. Your original identity is therefore equivalent
to 
$$
(ABD)=(ABE)+(ADF)\tag{1}
$$
Next, both the triangles $ABD$ and $ABC$ cut the parallelogram in 
two equal halves, so that (1) is equivalent to
$$
(ABC)=(ABE)+(ADF)\tag{2}
$$
Let $\lambda=\frac{BE}{EC}=\frac{CF}{FD}$.
Note that $EC=\frac{1}{1+\lambda}BC$. By Thales’ theorem, we deduce $(AEC)=\frac{1}{\lambda+1}(ABC)$. Note similarly that $CF=\frac{\lambda}{1+\lambda}CD$. By Thales’ theorem again, we deduce $(AFC)=\frac{\lambda}{\lambda+1}(ADC)$. 

It follows that 
$$
\begin{array}{lcl}
(ABE) &=& (ABC)-(AEC)=\frac{\lambda}{\lambda+1}(ABC)=\frac{\lambda}{\lambda+1}(ABC)\\
(ADF) &=& (ACD)-(ACF)=\frac{1}{\lambda+1}(ACD)=\frac{1}{\lambda+1}(ABC)\\
\end{array},
$$
 which finishes the proof.
