Two cevians divide a triangle into 4 parts. Calculate the area of the 4th part, given the other 3.

Question

Good day

Here is the question:

Connecting $AF$ and setting areas $\triangle ADF = x$ and $\triangle AFE = y$:

$\frac {9+x}{12} =\frac y{15}$

$\frac{15+y}{12} =\frac x9$ from the ratios of the sides to the areas.

I get the areas $x = 315$ and $y = 405$, though according to the image it seems impossible.

Am I missing something or is it correct?

Answer 1

Let $[ABC]$ be the area of $\triangle{ABC}$.

Let $[ADF]=x,[AEF]=y$. Then, from $$AD:DB=[ADF]:[DFB]=[ADC]:[CDB],$$ one has $$x:9=x+y+15:9+12\tag1$$

Also, from $$BF:FE=[BFA]:[FEA]=[BFC]:[CFE],$$ one has $$x+9:y=12:15\tag2$$

Solving $(1)(2)$ gives $$x=315,y=405.$$

Hence, $$[ADFE]=x+y=315+405=\color{red}{720}.$$