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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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A drawing would be helpful. –  Raskolnikov Aug 28 '12 at 16:07
    
If you have a drawing in front of you, @Nitesh, you'll see that it is impossible, without more data, that $\,EF=BF\,$ ,as this would render the triangles $\,\Delta AED\,,\,\,\Delta CBD\,$ congruent, and then $\,BD=ED\,$ , which is absurd as the latter segment is a subsegment of the former. –  DonAntonio Aug 28 '12 at 16:17
    
I added a diagram, but I don't see how AED and CBD have to be congruent if EF=BF? –  Nitesh Aug 28 '12 at 16:22
    
$$EF=FB\Longrightarrow \angle EBF=\angle BEF\,\,,\,\,\angle BEF=\angle AED\,\,(\text{vertex angles})$$ and since $\,AE=BC\,\,,\,\,AD=DC\,$ we get congruency –  DonAntonio Aug 28 '12 at 17:10
    
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. –  DonAntonio Aug 28 '12 at 17:13

1 Answer 1

enter image description here

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 $BC\geq AC/2$

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+1 Nice and simple. –  EuYu Aug 28 '12 at 19:29

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