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Difference between revisions of "Menelaus' Theorem"

(→‎Proof with Mass points)
(→‎Proof with Mass points)
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The mass at A is <math>m_{3}+m_{2}</math>
 
The mass at A is <math>m_{3}+m_{2}</math>
 
<cmath>AR\cdot (m_{3}+m_{2}) = RB \cdot m_{1} \Rightarrow \frac{AR}{RB} = \frac{m_{1}}{m_{3}+m_{2}} </cmath>
 
<cmath>AR\cdot (m_{3}+m_{2}) = RB \cdot m_{1} \Rightarrow \frac{AR}{RB} = \frac{m_{1}}{m_{3}+m_{2}} </cmath>
−
Multiplying them together, \fbox{<math>\frac{BP}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = \frac{{m_{2}}}{{m_{1}}} \cdot \frac{{m_{3}+m_{2}}}{{m_{2}}} \cdot \frac{{m_{1}}}{{m_{3}+m_{2}}} = 1</math>}
+
Multiplying them together, <math>\fbox{\frac{BP}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = \frac{{m_{2}}}{{m_{1}}} \cdot \frac{{m_{3}+m_{2}}}{{m_{2}}} \cdot \frac{{m_{1}}}{{m_{3}+m_{2}}} = 1}</math>
  
 
== Converse ==
 
== Converse ==

Revision as of 14:51, 13 December 2020

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

Proof with Mass points

Let's First define some points' masses.

$B_{m_{1}}$, $C_{m_{2}}$, and $Q_{m_{3}}$

By Mass Points: \[BP\cdot m_{1}=PC\cdot m_{2} \Rightarrow \frac{BP}{CP}=\frac{m_{2}}{m_{1}}\] \[\frac{QC}{QA}=\frac{AC+QA}{QA}=1+\frac{AC}{QA}=1+\frac{m_{3}}{m_{2}}=\frac{m_{2}}{m_{2}}+\frac{m_{3}}{m_{2}}=\frac{m_{3}+m_{2}}{m_{2}}\] The mass at A is $m_{3}+m_{2}$ \[AR\cdot (m_{3}+m_{2}) = RB \cdot m_{1} \Rightarrow \frac{AR}{RB} = \frac{m_{1}}{m_{3}+m_{2}}\] Multiplying them together, $\fbox{\frac{BP}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = \frac{{m_{2}}}{{m_{1}}} \cdot \frac{{m_{3}+m_{2}}}{{m_{2}}} \cdot \frac{{m_{1}}}{{m_{3}+m_{2}}} = 1}$ (Error compiling LaTeX. Unknown error_msg)

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

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