In the figure shown
We have $AD$ as median, $F$ is midpoint of $AD$, $BD=BF$. I wish to prove $AE=EF$. I tried to use Menlaus' theorem. but to no use. While doing the ratios taking two different triangles, I got two different values for the same ratio. For, in traingle $ABD$, we have the transversal $CE$. Using Menelaus', we get $$\frac{BC}{BD}\cdot\frac{DF}{AF}\cdot\frac{AE}{BE}=1$$ This implies $\frac{AE}{BE}=\frac1{2}\ldots\ldots(1)$. Now, in triangle $ABG$, we have the transversl $CE$. Again using Menelaus', we get $$\frac{AC}{GC}\cdot\frac{GF}{BF}\cdot\frac{BE}{AE}=1$$ Which implies that $\frac{AC}{GC}\cdot\frac{GF}{BF}=\frac1{2}\ldots\ldots(2)$. We take triangle $ACD$, with transversal $BG$ to get $$\frac{CB}{CD}\cdot\frac{DF}{AF}\cdot\frac{AG}{CG}=1$$ This implies $\frac{AG}{CG}=\frac1{2}\ldots\ldots(3)$. Using triangle $ACE$ with transversal $BG$ and Menelaus', we get: $$\frac{AB}{BE}\cdot\frac{EF}{CF}\cdot\frac{CG}{AG}=1$$ which implies that $\frac{EF}{CF}=\frac1{3}\ldots\ldots(4)$. Now, using the triangle $BCE$ with transversal $AD$, we have by Menelaus': $$\frac{AB}{AE}\cdot\frac{EF}{CF}\cdot\frac{CD}{BD}=1$$ which implies $\frac{EF}{CF}=\frac1{3}\ldots\ldots(5)$.
Using the equations $(1)$ through $(5)$, the only things I got are that $AE=\frac{AB}{3}$ and that $EF=\frac{CE}{4}$. Now, how do we proceed? Any hints? Thanks beforehand.
Much simpler approach:
Draw a line l from B parallel with AD. Extend CE and CA from E and A to intersect line l at H and I respectively.In triangle IBC , A is mid point of CI. H is mid point of BI, because F is mid point of AD , $AD||BI$ and $\triangle CAF \sim\triangle CIH$, ALso $\triangle CFD\sim \triangle CHB$. connect A to H. So in triangle IBC, AH connects mid points of sides IC and IB , so it is parallel with BC and quadrilateral AHBD is parallelogram, so we have:
$AH=BD=BF$
Hence quadrilateral AHBF is isosceles trapezoid and E is the intersection of it's diagonals.In this way triangles AHE and EBF are equal, therefore $AE=EF$
Note: I can not attach the figure I constructed due to my problem with this site. I can send you if I have an address.