Geometry with triangle ABC

Question

My question is...
I want to know your another solution.
or I want to know if my solution is appropriate.
and I’d appreciate some feedback on my work.

$E$ is midpoint of $\overline{BC}$.
$\overline{AD} : \overline{DE}$ = 4:3
Find the $\overline{AF} : \overline{FC}$.

Answer 1

Draw a paralel to $BF$ through $E$ and let it cuts $AC$ in $P$.

Then by Thales theorem we have $$FP:PC = BE:EC = 1:1$$

Again by Thales theorem we have $$AF:FP= AD:DE = 4:3$$

So, if $FP = 3x$ then $PC =3x$ and $AF = 4x$ so $$ AF:FC = 4x:6x = 2:3$$

Answer 2

Usually problems like this have one-line solution using Menelaus's theorem. For triangle $ACE$ and line $BDF$:

$$ \left|\frac{AF}{FC}\frac{CB}{BE}\frac{ED}{DA}\right| = 1,\qquad \frac{AF}{FC} = \frac{BE}{CB}\frac{DA}{ED} = \frac12 \frac{4}{3} = \frac 23\\ $$

Answer 3

Let me put together the two articles that answered my question.

Answer 4

There is a solution using the following

Theorem : If two triangles have their bases on a same line and have a common height, we have :

$$\frac{A_1}{A_2} = \frac{L_1}{L_2}$$

meaning that the ratio of their areas is the same as the ratio of the lengths of their bases. See for example http://www.madeiracityschools.org/userfiles/376/Classes/17052/Ratio%20of%20Areas-0.pdf.

Take a look at the following picture where the lower case letters indicate areas of the corresponding triangles.

We can write the following linear system where equations (1) to (4) are direct consequences of the upsaid theorem, the last equation (5) being a normalization (we are free to take the area unit we want ; taking $70$ is the result of an "afterthought", in order to provide integer values ) :

$$\begin{cases} (1)&x/z&=&4/3\\ (2)&y/v&=&4/3\\ (3)&v+z&=&w\\ (4)&v+w+y&=&x+z\\ (5)&v+w+x+y+z&=&70 \end{cases}$$

out of which one gets

$$(v,w,x,y,z) \ = \ (6, 21, 20, 8, 15).$$

The looked-for ratio is thus, using (backwards) the above theorem :

$$\frac{AF}{FC}=\frac{area(AEF)}{area(ECF)}=\frac{y+v}{w}=\frac{14}{21}=\frac{2}{3}.$$

Remark : Had we taken $1$ instead of $70$ in the RHS of equation (5), we would have obtained the same final result, but with unaesthetic fraction computations.