I am reading a 2006 book by Gerard A Venema titled "Exploring Advanced Euclidean Geometry With Geometer's Sketchpad." Chapter 9 deals with the Theorem of Menelaus. He covers the subject in more detail than I've seen anywhere else. He says that the theorem can be stated and proved even when one or more of the Menelaus Points are improper. An "improper" Menelaus Point is one that is identical with one of the vertices.
In the diagram below, \(\displaystyle ABC\) is an ordinary triangle, and L, M and N are Menelaus Points. A Menelaus Point is simply a point on one of the sidelines. For example, L is on the line that contains B and C. In the diagram it's shown as actually on the segment \(\displaystyle \overline {BL} \), but it could be anywhere else on \(\displaystyle BC\) and still be a Menelaus Point according to Dr. Venema. I have not seen the term "Menelaus Point" anywhere else.
The Theorem, as Dr. Venema states it, says that the three Menelaus Points are collinear if and only if \(\displaystyle s = - 1\), where
\(\displaystyle s = \frac{{\overline {AN} }}{{\overline {NB} }} \times \frac{{\overline {BL} }}{{\overline {LC} }} \times \frac{{\overline {CM} }}{{\overline {MA} }}\)
The books I've seen, other than Dr Venema's, deal only with cases where all the Menelaus Points are proper. Dr Venema also deals with the case of improper Menelaus Points. Here's roughly how he does it:
Assume, without loss of generality, that \(\displaystyle L = C\). Then, if the three Menelaus Points are to be collinear, either \(\displaystyle N = A\) or \(\displaystyle M = C\). So far, I'm on board with him. In either case, he says, \(\displaystyle s = - 1\). That's where he loses me.
Take the case where \(\displaystyle N = A\) and M might be something else. When I try to calculate s, I see no alternative but to write \(\displaystyle \overline {LC} = 0\)
and \(\displaystyle \overline {AN} = 0\), which leads to
\(\displaystyle s = \frac{0}{{\overline {NB} }} \times \frac{{\overline {BL} }}{0} \times \frac{{\overline {CM} }}{{\overline {MA} }}\)
and that makes no sense at all. I must be missing something simple. How do I get \(\displaystyle s = - 1\)?
Any help appreciated.
BTW, Dr Venema also says the theorem holds when the line NM is parallel to the line BC, because then they intersect at an "ideal point" that is infinitely distant. The ratio on line BC is then always -1. That sounds contrived to me, but I sort of follow it.
In the diagram below, \(\displaystyle ABC\) is an ordinary triangle, and L, M and N are Menelaus Points. A Menelaus Point is simply a point on one of the sidelines. For example, L is on the line that contains B and C. In the diagram it's shown as actually on the segment \(\displaystyle \overline {BL} \), but it could be anywhere else on \(\displaystyle BC\) and still be a Menelaus Point according to Dr. Venema. I have not seen the term "Menelaus Point" anywhere else.
The Theorem, as Dr. Venema states it, says that the three Menelaus Points are collinear if and only if \(\displaystyle s = - 1\), where
\(\displaystyle s = \frac{{\overline {AN} }}{{\overline {NB} }} \times \frac{{\overline {BL} }}{{\overline {LC} }} \times \frac{{\overline {CM} }}{{\overline {MA} }}\)
The books I've seen, other than Dr Venema's, deal only with cases where all the Menelaus Points are proper. Dr Venema also deals with the case of improper Menelaus Points. Here's roughly how he does it:
Assume, without loss of generality, that \(\displaystyle L = C\). Then, if the three Menelaus Points are to be collinear, either \(\displaystyle N = A\) or \(\displaystyle M = C\). So far, I'm on board with him. In either case, he says, \(\displaystyle s = - 1\). That's where he loses me.
Take the case where \(\displaystyle N = A\) and M might be something else. When I try to calculate s, I see no alternative but to write \(\displaystyle \overline {LC} = 0\)
and \(\displaystyle \overline {AN} = 0\), which leads to
\(\displaystyle s = \frac{0}{{\overline {NB} }} \times \frac{{\overline {BL} }}{0} \times \frac{{\overline {CM} }}{{\overline {MA} }}\)
and that makes no sense at all. I must be missing something simple. How do I get \(\displaystyle s = - 1\)?
Any help appreciated.
BTW, Dr Venema also says the theorem holds when the line NM is parallel to the line BC, because then they intersect at an "ideal point" that is infinitely distant. The ratio on line BC is then always -1. That sounds contrived to me, but I sort of follow it.