Attachment 20019 ABC is a isosceles triangle and DE is its central segment (I guess that's how you call it ); ADCF is parallelogram. DE = EF and AC is 10.

I was told to draw another segments from D to C and from B to F.

Attachment 20020

I have to find area of BDCF. I believe, that BDCF is a rectangle, even though my painting skills are very bad

So we have AC = 10, DE = 5, because DE= 2/AC. Find area of BDCF. My thoughts:

Hypotenuse DF of this BDCF $\displaystyle quadrangular = 10$. So we get a $\displaystyle triangle- DCF$ where C is straight angle. I thought DBE and EFC were parallel, so CF is aswell 5. The answer I got is 50, but it seems I missunderstood something.. ;\

Answer: $\displaystyle 25\sqrt{3}$