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Subject: "Menelaus Theorem"     Previous Topic | Next Topic
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Conferences The CTK Exchange College math Topic #43
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Sarah L Amstutz (Guest)
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Dec-11-00, 07:21 PM (EST)
 
"Menelaus Theorem"
 
   I just made my first visit to your web site the other day (I enjoyed it immensely!) and decided to do a project/presentation on the Menelaus Thm for my Conjecture and Proof course at Goshen College. This required an in-depth analysis of the proof(s) and in writing my paper, I noticed that the proofs on your site only proved "If the points are collinear, then AF/BF * BD/CD * CE/AE = 1," but the theorem is in an "if and only if" form. We learned in Conjecture and Proof class this semester that for an iff statement, the proof must prove, "If AF/BF * BD/CD * CE/AE = 1, then the points are collinear" also. Is the backward step so obvious that it wasn't needed and I missed something, or ...?

Thanks!

Sarah Amstutz


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alexb
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672 posts
Dec-11-00, 07:23 PM (EST)
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1. "RE: Menelaus Theorem"
In response to message #0
 
   Dear Sarah:

Thank you for the kind words.

Of course I should have mentioned this. The backward step if not obvious is standard. It's very much like that for Ceva's theorem:

https://www.cut-the-knot.com/Generalization/ceva.shtml.

Let there be three points such that AF/BF * BD/CD * CE/AE = 1 holds. Assume on the contrary that the points are not collinear. Pick up any two. Say D and E. Draw the line DE and find its intersection F' with AB. Then by the "forward" step AF'/BF' * BD/CD * CE/AE = 1. From which AF'/BF' = AF/BF. By subtracting 1
from both sides one gets AB/AF' = AB/AF, from which F' = F.

I have inserted this paragraph into the proof. Many thanks for pointing out this omission.

All the best,
Alexander Bogomolny


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