Difference between revisions of "Menelaus' Theorem"
(→Statement) |
Math.is.gud (talk | contribs) m (→Statement: deleted (-) in front of PC) |
||
(14 intermediate revisions by 10 users not shown) | |||
Line 1: | Line 1: | ||
− | |||
− | |||
'''Menelaus' Theorem''' deals with the [[collinearity]] of points on each of the three sides (extended when necessary) of a [[triangle]]. | '''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. | It is named for Menelaus of Alexandria. | ||
+ | |||
== Statement == | == Statement == | ||
− | |||
− | < | + | If line <math>PQ</math> intersecting <math>AB</math> on <math>\triangle ABC</math>, where <math>P</math> is on <math>BC</math>, <math>Q</math> is on the extension of <math>AC</math>, and <math>R</math> on the intersection of <math>PQ</math> and <math>AB</math>, then |
− | + | <cmath>\frac{PB}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = 1.</cmath> | |
− | |||
<center><asy> | <center><asy> | ||
+ | unitsize(16); | ||
defaultpen(fontsize(8)); | defaultpen(fontsize(8)); | ||
pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R; | pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R; | ||
Line 20: | Line 18: | ||
</asy></center> | </asy></center> | ||
− | == Proof == | + | Alternatively, when written with [[directed segment|directed segments]], the theorem becomes <math>BP\cdot CQ\cdot AR = PC\cdot QA\cdot RB</math>. |
+ | |||
+ | == Proofs == | ||
+ | |||
+ | ===Proof with Similar Triangles=== | ||
+ | |||
Draw a line parallel to <math>QP</math> through <math>A</math> to intersect <math>BC</math> at <math>K</math>: | Draw a line parallel to <math>QP</math> through <math>A</math> to intersect <math>BC</math> at <math>K</math>: | ||
<center><asy> | <center><asy> | ||
+ | unitsize(16); | ||
defaultpen(fontsize(8)); | defaultpen(fontsize(8)); | ||
pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R, K=(5.5,0); | pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R, K=(5.5,0); | ||
Line 35: | Line 39: | ||
<math>\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}</math> | <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{ | + | <math>\triangle QCP \sim \triangle ACK \implies \frac{QC}{QA}=\frac{CP}{KP}</math> |
Multiplying the two equalities together to eliminate the <math>PK</math> factor, we get: | Multiplying the two equalities together to eliminate the <math>PK</math> factor, we get: | ||
− | <math>\frac{AR}{RB}\cdot\frac{QC}{QA}= | + | <math>\frac{AR}{RB}\cdot\frac{QC}{QA}=\frac{CP}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{CP}=1</math> |
+ | |||
+ | ===Proof with [[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. | ||
Line 64: | Line 69: | ||
− | + | which yields, after simplification, | |
− | + | <cmath>-X\cdot (R-1)(P-1)+Y\cdot R(1-P)-Z\cdot PR = 0</cmath> | |
+ | <cmath>Z\cdot PR = -X\cdot (R-1)(P-1)+Y\cdot R(1-P).</cmath> | ||
+ | Plugging in the coordinates for <math>Q</math> yields <math>(Q-1)(R-1)(P-1) = QPR</math>. From <math>\frac{CP}{PB}=\frac{1-P}{P},</math> we have <cmath>P=\frac{(1-P)\cdot PB}{CP}.</cmath> Likewise, <cmath>R=\frac{(1-R)\cdot AR}{BR}</cmath> and <cmath>Q=\frac{(1-Q)\cdot QC}{QA}.</cmath> | ||
− | |||
+ | Substituting these values yields <cmath>(Q-1)(R-1)(P-1) = \frac{(1-Q)\cdot QC \cdot (1-P) \cdot PB \cdot (1-R) \cdot AR}{QA\cdot CP\cdot BR}</cmath> which simplifies to <math>QA\cdot CP \cdot BR = -QC \cdot AR \cdot PB.</math> | ||
− | + | QED | |
+ | == Converse == | ||
− | <math> | + | The converse of Menelaus' Statement is also true. If <math>\frac{BP}{PC} \cdot \frac{CQ}{QA} \cdot \frac{AR}{RB} = 1</math> in the below diagram, then <math>P, Q, R</math> are [[collinear]]. The converse is useful in proving that three points are collinear. |
+ | <center><asy> | ||
+ | unitsize(16); | ||
+ | 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> | ||
− | + | == See Also == | |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | == See | ||
* [[Ceva's Theorem]] | * [[Ceva's Theorem]] | ||
* [[Stewart's Theorem]] | * [[Stewart's Theorem]] |
Revision as of 19:34, 14 December 2019
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.
Contents
Statement
If line intersecting on , where is on , is on the extension of , and on the intersection of and , then
Alternatively, when written with directed segments, the theorem becomes .
Proofs
Proof with Similar Triangles
Draw a line parallel to through to intersect at :
Multiplying the two equalities together to eliminate the factor, we get:
Proof with 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:
Note that this says the following:
The line through and is given by:
which yields, after simplification,
Plugging in the coordinates for yields . From we have Likewise, and
Substituting these values yields which simplifies to
QED
Converse
The converse of Menelaus' Statement is also true. If in the below diagram, then are collinear. The converse is useful in proving that three points are collinear.