Difference between revisions of "Menelaus' Theorem"

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'''Menelaus' Theorem''' deals with the [[collinearity]] of points on each of the three sides (extended when necessary) of a [[triangle]].
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It is named for Menelaus of Alexandria.
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== Statement ==
 
== Statement ==
''(awaiting image)''
 
A necessary and sufficient condition for points D, E, F on the respective side lines BC, CA, AB of a triangle ABC to be collinear is that
 
<br><center><math>BD*CE*AF = -DC*EA*FB</math></center><br>
 
where all segments in the formula are [[directed segment]]s.
 
  
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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
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<cmath>\frac{PB}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = 1.</cmath>
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<center><asy>
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unitsize(16);
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defaultpen(fontsize(8));
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pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R;
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draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75);
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draw((7,6)--(6,8)--(4,0));
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R=intersectionpoint(A--B,Q--P);
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dot(A^^B^^C^^P^^Q^^R);
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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));
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</asy></center>
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Alternatively, when written with [[directed segment|directed segments]], the theorem becomes <math>BP\cdot CQ\cdot AR = PC\cdot QA\cdot RB</math>.
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== Proofs ==
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===Proof with Similar Triangles===
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Draw a line parallel to <math>QP</math> through <math>A</math> to intersect <math>BC</math> at <math>K</math>:
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<center><asy>
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unitsize(16);
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defaultpen(fontsize(8));
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pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R, K=(5.5,0);
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draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75);
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draw((7,6)--(6,8)--(4,0));
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draw(A--K, dashed);
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R=intersectionpoint(A--B,Q--P);
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dot(A^^B^^C^^P^^Q^^R^^K);
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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));
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label("K",K,(0,-1));
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</asy></center>
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<math>\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}</math>
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<math>\triangle QCP \sim \triangle ACK \implies \frac{QC}{QA}=\frac{CP}{KP}</math>
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Multiplying the two equalities together to eliminate the <math>PK</math> factor, we get:
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<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>
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===Proof with [[Barycentric coordinates]]===
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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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Suppose we give the points <math>P, Q, R</math> the following coordinates:
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<math>P: (0, P, 1-P)</math>
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<math>R: (R , 1-R, 0)</math>
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<math>Q: (1-Q ,0 , Q)</math>
  
== Example ==
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Note that this says the following:
''Does anyone have a good example? ''
 
  
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<math>\frac{CP}{PB}=\frac{1-P}{P}</math>
  
== See also ==
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<math>\frac{BR}{AR}=\frac{1-R}{R}</math>
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<math>\frac{QA}{QC}=\frac{1-Q}{Q}</math>
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The line through <math>R</math> and <math>P</math> is given by:
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<math>\begin{vmatrix} X & 0 & R \\ Y & P & 1-R\\ Z & 1-P & 0 \end{vmatrix} = 0</math>
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 +
 
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which yields, after simplification,
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<cmath>-X\cdot (R-1)(P-1)+Y\cdot R(1-P)-Z\cdot PR = 0</cmath>
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<cmath>Z\cdot PR = -X\cdot (R-1)(P-1)+Y\cdot R(1-P).</cmath>
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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>
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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>
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QED
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== Converse ==
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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.
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<center><asy>
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unitsize(16);
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defaultpen(fontsize(8));
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pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R;
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draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75);
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draw((7,6)--(6,8)--(4,0));
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R=intersectionpoint(A--B,Q--P);
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dot(A^^B^^C^^P^^Q^^R);
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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));
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</asy></center>
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== See Also ==
 
* [[Ceva's Theorem]]
 
* [[Ceva's Theorem]]
 
* [[Stewart's Theorem]]
 
* [[Stewart's Theorem]]
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[[Category:Theorems]]
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[[Category:Geometry]]

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.

Statement

If line $PQ$ intersecting $AB$ on $\triangle ABC$, where $P$ is on $BC$, $Q$ is on the extension of $AC$, and $R$ on the intersection of $PQ$ and $AB$, then \[\frac{PB}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = 1.\]

[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]

Alternatively, when written with directed segments, the theorem becomes $BP\cdot CQ\cdot AR = PC\cdot QA\cdot RB$.

Proofs

Proof with Similar Triangles

Draw a line parallel to $QP$ through $A$ to intersect $BC$ at $K$:

[asy] unitsize(16); 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]

$\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}$

$\triangle QCP \sim \triangle ACK \implies \frac{QC}{QA}=\frac{CP}{KP}$

Multiplying the two equalities together to eliminate the $PK$ factor, we get:

$\frac{AR}{RB}\cdot\frac{QC}{QA}=\frac{CP}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{CP}=1$

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 $P, Q, R$ the following coordinates:

$P: (0, P, 1-P)$

$R: (R , 1-R, 0)$

$Q: (1-Q ,0 , Q)$

Note that this says the following:

$\frac{CP}{PB}=\frac{1-P}{P}$

$\frac{BR}{AR}=\frac{1-R}{R}$

$\frac{QA}{QC}=\frac{1-Q}{Q}$

The line through $R$ and $P$ is given by: $\begin{vmatrix} X & 0 & R \\ Y & P & 1-R\\ Z & 1-P & 0 \end{vmatrix} = 0$


which yields, after simplification,

\[-X\cdot (R-1)(P-1)+Y\cdot R(1-P)-Z\cdot PR = 0\]

\[Z\cdot PR = -X\cdot (R-1)(P-1)+Y\cdot R(1-P).\]

Plugging in the coordinates for $Q$ yields $(Q-1)(R-1)(P-1) = QPR$. From $\frac{CP}{PB}=\frac{1-P}{P},$ we have \[P=\frac{(1-P)\cdot PB}{CP}.\] Likewise, \[R=\frac{(1-R)\cdot AR}{BR}\] and \[Q=\frac{(1-Q)\cdot QC}{QA}.\]


Substituting these values yields \[(Q-1)(R-1)(P-1) = \frac{(1-Q)\cdot QC \cdot (1-P) \cdot PB \cdot (1-R) \cdot AR}{QA\cdot CP\cdot BR}\] which simplifies to $QA\cdot CP \cdot BR = -QC \cdot AR \cdot PB.$

QED

Converse

The converse of Menelaus' Statement is also true. If $\frac{BP}{PC} \cdot \frac{CQ}{QA} \cdot \frac{AR}{RB} = 1$ in the below diagram, then $P, Q, R$ are collinear. The converse is useful in proving that three points are collinear.

[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]

See Also