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This question seems tricky and I frankly don't know how to start. I will be grateful if someone can provide a solution.

We have a triangle $ABC$ and there is a point $F$ on $BC$ such that $AF$ intersects the median $BD$ at $E$. If $AE=BC$ how do we prove that triangle $BEF$ is isosceles?

I think this has to do with ratios of sides, but I'm not getting any where. I drew a graph as accurately as I could and I am pretty confident that the objective of the problem is to somehow show that $EF$ equals $BF$, but I have no clue. Thanks!

http://postimage.org/image/gpcz69zxz/

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    Oops, wrong! We have side-side-angle, not side-angle-side! The above doesn't render those triangles congruent, indeed...unless $\,AD=DC\,$ is the longest side in each respective triangle.2012-08-28

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ABCG - parallelogram. BD is median $\Rightarrow$ B,D,E,G - collinear. We have $\angle AGE=\angle EBF$ and $\angle BEF=\angle AEG$

$\Delta ABC= \Delta ACG \Rightarrow AG=AE \Rightarrow $ $\angle AGE= \angle AEG \Rightarrow \angle EBF = \angle BEF$.

PS. F inside BC if only $BC\geq AC/2$