In the following figure: $AE = ED$ , $BD = DC$. Prove that $BE = \frac{EF}{3}$

Question

The question is : $AE = ED$ , $BD = DC$. Prove that $BE = \cfrac{EF}{3}$

Approach:

Realisations:

$AD$ is a median of triangle $ABC$.

$BE$ is a median of triangle $DBA$.

$E$ is a centroid.

I tried to prove these triangles similar, but I was unsuccessful as one criteria is always missing.

Please help me with this problem.

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How to do this using Ceva's Theorem?

Answer 1

Denote areas as [.] and establish the ratio $$1=\frac{AE}{ED}= \frac{[ABF]}{[DBF]} = \frac{\frac{AF}{AC}[ABC]}{\frac12 (1-\frac{AF}{AC})[ABC]}\implies \frac{AF}{AC}=\frac13 $$ Then $$ \frac{BE}{EF}= \frac{[ABD]}{[AFD]} = \frac{\frac12[ABC]}{\frac12 \frac{AF}{AC}[ABC]}=\frac{AC}{AF}=3 $$

Answer 2

Apply Menelaus' Theorem following B-C-A-D to see:

$$\frac{BD}{BC} \cdot \frac{FC}{AF}\cdot \frac{AE}{ED} =1 \implies \frac{AF}{FC}=\frac{1}{2}$$

Apply Menelaus' Theorem again, this time following A-C-B-F to see:

$$\frac{AF}{AC} \cdot \frac{DC}{BD}\cdot \frac{BE}{EF}=1 \implies \frac{BE}{EF}=3$$