In the given triangle with two intersecting cevians, find the area of the shaded quadrilateral. As title suggests, the question is to find the area of the shaded quadrilateral inside a triangle formed via two intersecting cevians, given the area of $3$ smaller triangles with areas $3$, $6$ and $9$ respectively.

I discovered this puzzle online posted on a language-learning app, and I found it quite interesting. The asker claimed that it was a grade 9 problem. I did attempt it in several ways, but a lot of the attempts did not lead anywhere. I will post my successful attempt as an answer below, please do let me know if it is correct, if something can be improved or if the answer may be wrong (the asker did not reveal the answer). And please share your own attempts too!
 A: Let us denote by $X$ the area to be computed, and let us use letters for the relevant vertices as in the following picture:

Then applying Menelaus in $\Delta EFB$ w.r.t. the line $CDA$ we get:
$$
1=
\frac{CE}{CF}\cdot
\frac{DF}{DB}\cdot
\frac{AB}{AE}
=
\frac{6+9}{9}\cdot
\frac{3}{3+9}\cdot
\frac{X+3+6+9}{X+3}\ ,
$$
which gives a first order equation in $X$ with unique solution $X=54/7$.
A: Here is my attempt at it:

1.) First, let's label the triangle $\triangle ABC$, the points at which the cevians touch the triangle as $D$ and $E$ and the point of intersection of the cevians as $F$. Draw $AF$ such that it divides Quadrilateral $AEFD$ into two separate triangles with areas $x$ and $y$ respectively.
2.) Now, recall the lemma that the ratio of the areas of two triangles divided by any line, which share the same height, is equal to the ratio of their bases (this can easily be proven). Therefore, we can claim that:
$$\frac{AD}{DC}=\frac{x}{3}$$ and,
$$\frac{AE}{EB}=\frac{y}{6}$$
With this in mind, we can extend this claim and say that:
$$\frac{x}{3}=\frac{x+y+6}{12}$$
$$\frac{y}{6}=\frac{x+y+3}{15}$$
Now we have a system of equations with two variables that we can easily solve with many different methods. I will leave it up to the readers to solve it with a method of their own choice, but to keep this post from getting unnecessarily lengthy, I will reveal that $x=\frac{24}{7}$ and $y=\frac{30}{7}$
Therefore the area of the shaded quadrilateral $AEFD$ is:
$$x+y=\frac{24}{7}+\frac{30}{7}=\frac{54}{7}$$
