Difference between revisions of "Menelaus' Theorem"
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It is named for Menelaus of Alexandria. | It is named for Menelaus of Alexandria. | ||
== Statement == | == Statement == | ||
− | A necessary and sufficient condition for points <math> | + | A necessary and sufficient condition for points <math>P, Q, R</math> on the respective sides <math>BC, CA, AB</math> (or their extensions) of a triangle <math>ABC</math> to be collinear is that |
− | <center><math> | + | <center><math>BP\cdot CQ\cdot AR = -PC\cdot QA\cdot RB</math></center> |
where all segments in the formula are [[directed segment]]s. | where all segments in the formula are [[directed segment]]s. | ||
− | + | <center><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></center> | ||
+ | == Proof == | ||
+ | Draw a line parallel to <math>QP</math> through <math>A</math> to intersect <math>BC</math> at <math>K</math>: | ||
+ | <center><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></center> | ||
+ | <math>\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}</math> | ||
+ | |||
+ | <math>\triangle QCP \sim \triangle ACK \implies \frac{QC}{QA}=\frac{PC}{PK}</math> | ||
+ | |||
+ | Multiplying the two equalities together to eliminate the <math>PK</math> factor, we get: | ||
+ | |||
+ | <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> | ||
== See also == | == See also == | ||
* [[Ceva's Theorem]] | * [[Ceva's Theorem]] |
Revision as of 00:01, 11 March 2008
This article is a stub. Help us out by expanding it.
Menelaus' 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: