I am completely at a loss as to how to approach this problem
it's due in about 18 hours, so I'll be working on it only a little bit more, as I also have to sleep sometime tonight
all I've been able to deduce is that [ABF]=[AFC], [BDF]=[FDC]
[AFC]=[AED]+[DEF]+[FDC]
[AFC]=[AED]+1+[BDF]
[ABE]=[ABF]-1
and from there it's just substitution of what I already know and no elimination
and I know we don't provide full solutions, I had some thanks to my name in the past but they seem to be gone since the forum software migration (I spent a long time away)
Mais tabarnak Monsieur Pellerin, c'est quoi cette hostie de torture ?!
Well, a bit too much for my humble capabilities (plus I hate anything involving Heron's formula!);
so I set up a program looking at AB=AC in 1/4 increments, similarly with BC, and got this:
AB=AC=65 and BC=1/4 (yes, 1/4):
results in areaDEF = .99999815.... (darn close to 1)
and areaCDF = .06249988... (suspiciously close to 5/8)
and AD being 65-1 = 64
So seems you'll get an awfully long skinny isosceles ABC.
Probably won't help you much...but what the heck: any reply is better than none
using menalaus
I got AD = 3/2
then menalaus again BE/ED = 5/3
then the base*height ratio thing (forget the name)
I get 5/3 = BE/ED = [BEF]/[DEF] = [BEF]
and then [DFC]=[BDF]=[DEF]+[BEF] = 1+5/3 = 8/3
your solution didn't make sense since it would make [BFD] less than 1, but [BFD] contains [DEF]=1
hurray for sleeping on it then trying again with fresh insight