Let $ABCD$ be a parallelogram. By a dot $ x$ on the $\overline {AB}$ with $\overline {AX} <\overline{XB}$, draw a line parallel to $\overline{BC}$. This line, along with the diagonals and sides of the parallelogram, will determine $3$ quads. Knowing that the sum of the areas of these quads is as large as possible, calculate the ratio $\frac{\overline{AX}} {\overline{AB}}$
I tried to explain the reasons between perimeters and harmonic beams, but it didn't work on this issue.
Let the area of ABCD be 1 and, with the ratios of the similar triangles, calculate the areas of the triangle $a$, $b$ and $c$ as,
$$A_a = A_b= \frac{AX^2}{AB^2} A_{ACB}= \frac12 r^2$$
$$A_c = \frac{(\frac{AB}{2}-AX)^2}{(\frac{AB}{2})^2}A_{ADO} = \frac14 (1-2r)^2$$
where $r=\frac{AX}{AB}$, $A_{ACB}=\frac12$ and $A_{ADO} = \frac14$. The sum of $a$, $b$ and $c$ is then,
$$A(r) = r^2+\frac14(1-2r)^2$$
Take the derivative of $A(r)$ with respect to $r$ and set it to zero to minimize $A$, hence maximizing the sum of the three quads. We obtain the ratio,
$$r=\frac{AX}{AB}=\frac14$$