Date: 01/25/99 at 11:54:49 From: Wanwipa Subject: Using Menelaus' theorem to prove this problem I want to know how to prove this problem by using Menelaus' theorem: A straight line intersects sides AB, BC and the extension of side AC of a triangle ABC at points D, E and F respectively. Prove that the midpoints of the line segments DC, AE and BF lies on a straight line. Thank you Wanwipa.
Date: 01/26/99 at 09:27:17 From: Doctor Floor Subject: Re: Using Menelaus' theorem to prove this problem Hi Wanwipa, Thanks for your question. First, for your reference, here is a diagram of the set-up: Note that in this picture, points D, E and F are not the same as in your question. Gauss' theorem (1810) says: The midpoints of segments AD, BE and CF are collinear. In the figure they are marked G, H and I. Now apply Menelaus' theorem on triangle ABC and line DE, to find: AF BE CD -- * -- * -- = -1 FB EC DA Consider triangle JKL, made out of the side midpoints of triangle ABC. You can see that JI = 0.5 BE and IL = 0.5 EC, so BE/EC = JI/IL. In the same way you find: AF/FB = LH/HK and CD/DA = KG/GJ. So we have: LH JI KG -- * -- * -- = -1 HK IL GJ So, again applying Menelaus' theorem, this must mean that G, H, and I are collinear. That is the Gaussian line. If you have a math question again, please send it to Dr. Math. If necessary, don't hesitate to write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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