Suppose $AB/AF=CB/CD=3$ Prove that $E$ is the midpoint of $\overline{AC}$

Question

Suppose $AB/AF=CB/CD=3$ and the three cevians $\overline{AD},\overline{BE},\overline{CF}$ are concurrent. Prove that $E$ is the midpoint of $\overline{AC}$

I know by cevas theorem that since the cevians are concurrent then $AF/FB \cdot BD/DB \cdot AE/EC=1$ So I just need to show that $AF/FB \cdot BD/DB=1$. However just trying to manipulate the given it didn't come out nicely.

Answer 1

Since $AB=3AF$, $FB=AB-AF=2AF$.

Since $CB=3CD$, $BD=CB-CD=2CD$.

$$\frac{AF}{FB}=\frac{1}{2}$$

$$\frac{BD}{DC}=2$$

By Ceva's Theorem,

$$\frac{AF}{FB}\times\frac{BD}{DC}\times\frac{CE}{EA}=1$$

So, $CE=EA$

Answer 2

Hint 1: Have you heard of Ceva theorem? It is useful here.

Hint 2: You can easily solve it with vectors.

Hint 3: You can solve it with Thales theorem and Sylvester theorem.

Hint 4:

You can solve it with area. Remember that $A/A' = a/a'$ if $A,A'$ are areas of triangles with bases $a,a'$ and the same height.