What is Menelaus' Theorem?
Date: 11/15/98 at 16:00:25 From: Bob Subject: Menelaus' Theorem I need to write a paper on the Menelaus' Theorem. I need some information on it or references for it. Thanks.
Date: 11/16/98 at 10:44:43 From: Doctor Floor Subject: Re: Menelaus' Theorem Hi Bob, Thanks for your question. Let me try to tell you a little bit about Menelaus' Theorem. Let ABC be a triangle and l a line that intersects the (produced) sides of ABC in P (on BC), Q (on AC), and R (on AB). Make a line through C parallel to AB, and let the intersection with l be C'. Observe that triangles CPC' and BPR are similar. So CC':RB = PC:BP, and thus: CC' BP --- * -- = 1  RB PC Note also that triangles RAQ and C'CQ are similar. So AR:CC' = AQ:CQ, and thus: AR CQ --- * -- = 1  CC' AQ Multiplying  and  gives: CC' BP AR CQ --- * -- * --- * -- = 1 RB PC CC' AQ Cancelling CC' from numerator and denominator gives: BP AR CQ -- * -- * -- = 1 PC RB AQ From the observation that AQ = -QA we find: BP AR CQ -- * -- * -- = -1  PC RB QA This identity , for three collinear points P, Q and R on the sides of a triangle ABC, is Menelaus' Theorem. Menelaus proved it in around 100 BC. The converse of the theorem is also true: If three points P, Q and R on the sides of triangle ABC fit in , then they are collinear. A famous theorem that depends on Menelaus' Theorem is Desargues' Theorem (1639): Let /\ denote intersection. Let ABC and A'B'C' be two triangles. If the lines AA', BB', and CC' concur in one point, then the intersection points AB/\A'B', BC/\B'C' and AC/\A'C' are collinear. Here also the converse is true: if the three mentioned intersection points are collinear, then AA', BB', and CC' concur in one point. A biography of Menelaus can be found in the MacTutor Math History archives at: http://www-history.mcs.st-and.ac.uk/history/Mathematicians/Menelaus.html I hope this helps. If you have a math question again, please send it to Dr. Math. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
Date: 05/03/2001 at 18:50:41 From: Krish Subject: A question on the Menelaus theorem. I just discovered this site, and my, am I ever amazed. You guys are awesome. Thought you should know. Well, my question has to do with a statement that was written in the Menelaus theorem answer that you gave. The theorem concludes as follows: This identity , for three collinear points P, Q and R on the sides of a triangle ABC, is Menelaus' Theorem. This is fine except for one little problem: in the drawing, R isn't on the triangle ABC as such. Yes, if you extend the segment AB, R intersects it, but not on the segment itself. Can R then be considered to be "on the side of the triangle"? As well, just before concluding, you state that because AQ = -QA, the product of the left side gives -1. Why is AQ equal to negative QA if both signify the lengths of segments? I was always under the impression that lengths are positive. One more quick question: I have been studying for an international exam that requires that I know Apollonius' theorem. I'm certain Apollonius had many theorems, so do you know any particular one(s) that he was most famous for? Thanks for your time. KRish
Date: 05/04/2001 at 13:57:16 From: Doctor Schwa Subject: Re: A question on the Menelaus theorem. Thanks! Before the dot-com explosion, this kind of site was what the promise of the Internet was all about: putting together information in a useful way. I'm proud to be a volunteer producing one (very tiny!) fraction of what Dr. Math does, which in turn is but a fraction of what the Math Forum has to offer... >This is fine except for one little problem: in the drawing, R isn't >on the triangle ABC as such. Yes, if you extend the segment AB, R >intersects it, but not on the segment itself. Can R then be >considered to be "on the side of the triangle"? As you notice, it depends on how you define "the side of the triangle." It's very useful sometimes to think of a triangle as being made of three lines instead of three segments. This theorem is one example. Also, when you start analyzing "excircles," if you draw the lines, you realize that each angle of the triangle has two different angle bisectors (at right angles to each other), and intersecting them in various combinations can give you other circles tangent to the three (extended) sides of the triangle, just like the incircle. So, when we start talking Menelaus, we use the lines instead of the segments. If you're interested in this sort of thing, I *highly* recommend Coxeter's excellent little book, _Geometry Revisited_. Expect to spend about an hour per page to understand everything ... but there sure is a lot of geometry to be learned there. >As well, just before concluding, you state that because AQ = -QA, the >product of the left side gives -1. Why is AQ equal to negative QA if >both signify the lengths of segments? I was always under the >impression that lengths are positive. Again, this is a convention adopted for this particular type of problem. Probably we didn't make that clear enough in this particular discussion. Coxeter's book does a careful job of explaining when we call a distance positive or negative, and why that convention is so useful in this type of problem. >One more quick question: I have been studying for an international >exam that requires that I know Apollonius' theorem. I'm certain >Apollonius had many theorems, so do you know any particular one(s) >that he was most famous for? Thanks for your time. He's most famous for a construction, not a theorem: given three circles, construct a circle tangent to all three of them. Also the related problems (where some of the circles are replaced by lines). He's famous for work on conic sections in general, and a good discussion of his life and work is in the MacTutor Math History archives at St. Andrews, but it doesn't name any particular item "Apollonius' theorem." http://www-history.mcs.st-and.ac.uk/history/Mathematicians/Apollonius.html Stewart's theorem is often also credited to Apollonius as well. In the figure in our archives for "What is Menelaus' Theorem?" at http://mathforum.org/dr.math/problems/sonsino5.18.96.html it says na^2 + mb^2 = c(d^2 + mn) but I always remember it as "bomb + sink = dad + man," or, in other words, bmb + cnc = dad + man If you switch the side labels "a" and "c" in the picture on that page, and then replace things like mb^2 = mbb = bmb, then you'll get the formula for Stewart's (aka Apollonius') theorem in my alternate, easier-to-remember form. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum