# Thread: area proof, maybe menelaus

1. ## area proof, maybe menelaus

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

2. What do you mean by $\displaystyle \textrm{[D\,E\,F]}$ and $\displaystyle \textrm{[D\,F\,C]}$?

3. Area of triangle DEF and triangle DFC, respectively

4. And what work have you done yourself so far? It's not in MHF's policy to knowingly complete work that will be graded - only to give a nudge in the right direction...

5. 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)

6. 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.

7. using menalaus
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

8. Originally Posted by jbpellerin
using menalaus I got AD = 3/2 .....
your solution didn't make sense since it would make [BFD] less than 1, but [BFD] contains [DEF]=1
Yep; my bad; at least I got to meet Mr. Menalaus!