Parallelogram and Areas

Question

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 $(AKL)=(BEK)+(DLF)$

Answer 1

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.