Não importa o quanto as Áreas se escondam, ainda as encontraremos // No matter how much the Areas hide, we will still find them

           Based on a LinkedIn Post

Then I found a similar one

I thought of a general formula, resulting in the following:

          PROBLEM In the parallelogram $ABCD$ in the figure below, we denote the

                        areas of the triangles $ABM,\;KLM$ by $\textbf{S}_1,\; \textbf{S}_2$ and the areas of the

                        quadrilaterals $ADLM,\;BCKM$ by $\textbf{S}_3,\;\textbf{S}_4$. The following                                              relationship holds:

$$\textbf{S}_3+\textbf{S}_4=\textbf{S}_1-\textbf{S}_2+2 \cdot \sqrt{\textbf{S}_1 \cdot \textbf{S}_2} \tag{1}$$

          Solution CiP

               I added additional lines and notations to the figure, compared to those mentioned in the statement: $$EF \parallel AB,DC,\;EF \ni M,\;\;PQ\perp AB,DC,\;PQ \ni M.$$

          It is known that the ratio of the areas of two similar triangles

$\frac{\textbf{S}_2}{\textbf{S}_1}=\left ( \frac{MQ}{MP} \right )^2\;\;\Rightarrow\;\frac{MQ}{MP}=\sqrt{\frac{\textbf{S}_2}{\textbf{S}_1}} \tag{2}$

We also have $\frac{\textbf{S}_{CEFD}}{\textbf{S}_{ABEF}}=\frac{EF \cdot MQ}{EF \cdot MP}=\frac{MQ}{MP}\underset{(2)}{=}\sqrt{\frac{\textbf{S}_2}{\textbf{S}_1}}\;\;\;\Rightarrow \;\textbf{S}_{CEFD}=\textbf{S}_{ABEF} \cdot \sqrt{\frac{\textbf{S}_2}{\textbf{S}_1}} \tag{3}$

     On the other hand $\textbf{S}_{ABEF}=AB \cdot MP=2\cdot \textbf{S}_{ABM}=2\cdot\textbf{S}_1\;\;\overset{(3)}{\Rightarrow}\;\textbf{S}_{CEFD}=2\cdot \sqrt{\textbf{S}_1\cdot \textbf{S}_2}$

and      $\textbf{S}_{ABCD}=\textbf{S}_{ABEF}+\textbf{S}_{CEFD}=2\cdot \textbf{S}_1+2\cdot \sqrt{\textbf{S}_1 \cdot \textbf{S}_2}\;\;\;\Leftrightarrow$

$\Leftrightarrow \textbf{S}_1+\textbf{S}_2+\textbf{S}_3+\textbf{S}_4=2\cdot \textbf{S}_1+2\cdot \sqrt{\textbf{S}_1 \cdot \textbf{S}_2}\;\;\;\Rightarrow \textbf{S}_3+\textbf{S}_4=\textbf{S}_1-\textbf{S}_2+2\cdot \sqrt{\textbf{S}_1 \cdot \textbf{S}_2}$

QED $\blacksquare$