Proving Triangle AEF = Triangle FBD: A Puzzling Problem

[ruud]

I can't figure out this problem

Lets take triangle ABC
A


B C

a median (D) goes from A to the midpoint of BC
a median (E) goes from B to the midpoint of AC

Prove that triangle AEF = triangle FBD

Since a median divides up a triangle you know that
triangle ABD = triangle ADC
triangle ABE = triangle EBC

Here is where I'm getting stuck could someone please tell me the next step or two?

 

[ruud]

I forgot to mention F is the point where the two medians cross

 

[paul11273]

Well, I am not sure what form your answer is supposed to take, but how about this...

You know that
ABD = ADC

Then when you bisect them it follows that:
(1/2)ABD = (1/2)ADC

Since
BDF = (1/2)ABD and AEF = (1/2)ADC

Then
BDF = AEF

I hope this helps. Let me know.