A geometry problem - might be an open one

[ins-]

Let $S_1$, $S_2$, $S_3$, $S_4$, $S_5$, $S_6$ are the areas of the six triangles, formed by a triangle's cevians. Find $S_2$, $S_4$, $S_6$ in terms of $S_1$, $S_3$, $S_5$.

[DeBill]

Нумеруя треугольники по кругу, начиная от вершины, цифрами 1...6, получим три равенства, типа
$\frac{1}{2} = \frac{5+6}{3+4}$... Три гиперболических параболоида. Не решается...
Есть, конечно, два красивых соотношения:
$1\cdot 3\cdot 5 = 2\cdot 4\cdot 6$, и
$1\cdot 3+3\cdot 5+5\cdot 1 = 2\cdot 4 + 4\cdot 6 + 6\cdot 2$, да толку то...

[ins-]

$\frac{AP}{PA_1}+\frac{BP}{PB_1}+\frac{CP}{PC_1}+2=\frac{AP.BP.CP}{PA_1.PB_1.PC_1}$ if the triangle is $ABC$ and cevians are $AA_1$, $BB_1$, $CC_1$ with intersection point $P$ from this equality one more relationship between the areas can be found. There are two more similar relations between the cevians. Thus we can find 3-4 equations with 6 variables. The 3 variables are defined in terms of other 3 but I suppose it is not an easy problem, if even solvable, because I found this one: http://www.qbyte.org/puzzles/p002s.html, one similar Sangaku problem and https://en.wikipedia.org/wiki/Routh%27s_theorem, but for this one - nothing. If we find the area of $ABC$ in terms of $S_1$, $S_3$, $S_5$ the problem is solved. Another open problem is - if $S_1=S_3=S_5$ is the triangle $ABC$ equilateral?

[ins-]

The answer of last question is obvious - if the three cevians are medians - $S_1=...=S_6$, so the triangle is not always equilateral.