Nonparallel lines in triangles My kid had this geometry problem:

Given triangle $ABC$ with midpoint $E$ of $\overline{AB}$ and a point $D$ on $\overline{BC}$. $\overline{AD}$ and $\overline{CE}$ meet at point $F$. $AF=3DF$. Prove that $BD=2DC$.

Here's a rough sketch for convenience:

I wanted to argue as follows, but it uses theorems that my kid's class has not encountered. Nor indeed have I, which is why I'm seeking your feedback.

As $CE$ projects onto $BC$ in the cone from $A$, it reaches half the height of the cone at point $E$ but only a fourth of the cone's height at point $F$. Thus, $F$ is the midpoint of $CE$.
Likewise, as $AD$ projects onto $AB$ in the cone from $C$, it reaches half the height of the cone at point $F$, three-fourths of the way through, and thus $\frac12\frac43$ the height of the cone at $D$, which is what we wanted to show.

Does that work? (I'm asking not whether that proof is based on stuff a high-schooler may have learned but whether it works at all. Though if it or some special case is learned in high-school geometry, all the better.) If so, what is the theorem I used called, and what's a correct statement of it?
 A: A short proof using a standard result in projective geometry is as follows:
We apply the theorem of Menelaus of Alexandria, wiki, Menelaus' Theorem,

for the triangle $\Delta ABD$ using the line $EFC$ that intersects the sides of the triangle in the mentioned points. From the relation
$$
1=
\underbrace{\frac {EA}{EB}}_{=1}\cdot
\frac {CB}{CD}\cdot
\underbrace{\frac {FD}{FA}}_{=1/3}
$$
we obtain immediately the needed relation in the form $CB=3CD$.

The given proof is using some idea that may be used to give a proof of the Menelaus' Theorem in its full generality.
To understand the given proof as it is, in a special case, let us project $A;E,F$ onto $BC$ respectively in points denoted by $A';E', F'$.

Then the text wants to argue as follows. We have 
$$
\begin{aligned}
\frac{EE'}{AA'} &= \frac {BE}{BA} =\frac 12 \ ,\\
\frac{FF'}{AA'} &= \frac {DF}{DA} =\frac 14 \ ,\\[3mm]
&\qquad\text{ and from here}\\
\frac{CF}{CE}=
\frac{FF'}{EE'} &= 
\frac{FF'}{AA'}\cdot\frac{AA'}{EE'}
= \frac 14\cdot\frac 21 =\frac 24=\frac 12 \ .
\end{aligned}
$$
So $F$ is indeed the mid point of $CE$. (I did the above as detailed as possible.)
Now the suggested solution wants to project further onto $AB$. It is hard to draw and insert here, so please take a sheet of paper, draw the projections $C^*; D^*,F^*$ of the points $C;D,F$ onto $AB$. Then we have in a row this time:
$$
\frac{BD}{BC} =
\frac{DD^*}{CC^*} =
\frac{DD^*}{FF^*} \cdot
\frac{FF^*}{CC^*} =
\frac{AD}{AF} \cdot
\frac{EF}{EC} 
=\frac 43\cdot\frac 12
=\frac 23
\ .
$$

Subjective conclusion: Kids always get problems that can be solved easily by known results, but the proposer knows they do not know that result. The proof involves ideas from the proof of the known result. Didactically, this may serve as a wonderful introduction to the beautiful proof of the important and beautiful known result. Ideally, the kids will find themselves the proof based on the baby example. Practically, this makes kids stay at distance from mathematics. Moreover, such discussions always occur, we use a situation at half distance to an established result, to understand and explain to the kids what the example wanted and what the theorem wanted, will make the kids understand better where the teaching methods fail. In this perfect world of recognized mathematical beauty they are strong enough to accept the situation and put themselves in the position to know both, the ad-hoc solution in the simpler case, and the theorem with proof.
A: Your proof works. 
What is reassuring, especially for high-schoolers, is that there is no advanced theorem employed. Your argument is entirely based on the property of similar triangles.
For each ratio you derived, there is a particular pair of right triangles that you can identify for your argument. For example, the first ratio you obtained, 
$$\frac{h_E}{h_A}=\frac{BE}{BA}=\frac{1}{2}$$
i.e. the height $h_E$ of the point E is half of the height $h_A$ of the point A is to be based on the fact that the triangles BYE and BXA are similar, where Y and X are the two intersects of the heights with the base BC, respectively. The above ratio for heights results exclusively from the property of the similar right triangles BYE and BXA. 
Similar pairs of right triangles can be identified for your derivation of all the other ratios needed for the proof.
A: This problem is essentially an application of “area is proportional to base if the altitudes are the same”.

In the figure, let [BFD] = 2h for some $h \ne 0$. 
Then, [FBE] =[FAE] = 6h/2 = 3h
Let [FDC] = k for some $k \ne 0$. Then, [FAC] = 3k
From [CBE] = [CAE], we have 3h + 2h + k = 3h + 3k. Therefore, h = k. Result follows.
