Difference between revisions of "Menelaus' Theorem"
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<math>\frac{AR}{RB}\cdot\frac{QC}{QA}=-\frac{PC}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{PC}=-1</math> | <math>\frac{AR}{RB}\cdot\frac{QC}{QA}=-\frac{PC}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{PC}=-1</math> | ||
− | ==Proof Using [[Barycentric | + | ==Proof Using [[Barycentric coordinates]]== |
Disclaimer: This proof is not nearly as elegant as the above one. It uses a bash-type approach, as [[barycentric coordinate]] proofs tend to be. | Disclaimer: This proof is not nearly as elegant as the above one. It uses a bash-type approach, as [[barycentric coordinate]] proofs tend to be. | ||
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QED | QED | ||
+ | |||
== See also == | == See also == | ||
* [[Ceva's Theorem]] | * [[Ceva's Theorem]] |
Revision as of 00:31, 29 May 2013
This article is a stub. Help us out by expanding it.
Menelaus's Theorem deals with the collinearity of points on each of the three sides (extended when necessary) of a triangle. It is named for Menelaus of Alexandria.
Statement
A necessary and sufficient condition for points on the respective sides
(or their extensions) of a triangle
to be collinear is that
![$BP\cdot CQ\cdot AR = -PC\cdot QA\cdot RB$](http://latex.artofproblemsolving.com/9/7/4/974c07d36b8fe13f0b58f5d2ac02809f9c78a1ec.png)
where all segments in the formula are directed segments.
![[asy] defaultpen(fontsize(8)); pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R; draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75); draw((7,6)--(6,8)--(4,0)); R=intersectionpoint(A--B,Q--P); dot(A^^B^^C^^P^^Q^^R); label("A",A,(1,1));label("B",B,(-1,0));label("C",C,(1,0));label("P",P,(0,-1));label("Q",Q,(1,0));label("R",R,(-1,1)); [/asy]](http://latex.artofproblemsolving.com/8/6/a/86a254a29864a11196f5bbd900856e311e41be6b.png)
Proof
Draw a line parallel to through
to intersect
at
:
![[asy] defaultpen(fontsize(8)); pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R, K=(5.5,0); draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75); draw((7,6)--(6,8)--(4,0)); draw(A--K, dashed); R=intersectionpoint(A--B,Q--P); dot(A^^B^^C^^P^^Q^^R^^K); label("A",A,(1,1));label("B",B,(-1,0));label("C",C,(1,0));label("P",P,(0,-1));label("Q",Q,(1,0));label("R",R,(-1,1)); label("K",K,(0,-1)); [/asy]](http://latex.artofproblemsolving.com/f/b/d/fbd7bf9ba4bcb21c083f9f6aef5d813523b36c8f.png)
Multiplying the two equalities together to eliminate the factor, we get:
Proof Using Barycentric coordinates
Disclaimer: This proof is not nearly as elegant as the above one. It uses a bash-type approach, as barycentric coordinate proofs tend to be.
Suppose we give the points the following coordinates:
The line through and
is given by:
![$\begin{vmatrix} X & 0 & R \\ Y & P & 1-R\\ Z & 1-P & 0 \end{vmatrix} = 0$](http://latex.artofproblemsolving.com/1/4/d/14d4e372cdb043b6d23a6685f6f2c569dc27cc16.png)
Which yields, after simplification,
$Z\cdotPR = -X\cdot(R-1)(P-1)+Y\cdotR(1-P)$ (Error compiling LaTeX. Unknown error_msg)
Plugging in the coordinates for yields:
QED