Difference between revisions of "Menelaus' Theorem"
Math.is.gud (talk | contribs) m (→Statement: deleted (-) in front of PC) |
(→Proofs) |
||
Line 81: | Line 81: | ||
QED | QED | ||
+ | |||
+ | ===Proof with Mass Points=== | ||
+ | Let's First define some points' masses. | ||
+ | |||
+ | <math>B_{m_{1}}</math>, <math>C_{m_{2}}</math>, and <math>Q_{m_{3}}</math> | ||
+ | |||
+ | By Mass Points: | ||
+ | <cmath>BP\cdot m_{1}=PC\cdot m_{2} \Rightarrow \frac{BP}{CP}=\frac{m_{2}}{m_{1}}</cmath> | ||
+ | <cmath>\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}}</cmath> | ||
+ | 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> | ||
+ | Multiplying them together, <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> | ||
== Converse == | == Converse == |
Revision as of 15:47, 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.
Contents
Statement
If line intersecting
on
, where
is on
,
is on the extension of
, and
on the intersection of
and
, then
![[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]](http://latex.artofproblemsolving.com/8/7/e/87e7bd56f19d2bc7aa2f8aa9681e95b0d546ef76.png)
Alternatively, when written with directed segments, the theorem becomes .
Proofs
Proof with Similar Triangles
Draw a line parallel to through
to intersect
at
:
![[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]](http://latex.artofproblemsolving.com/2/1/a/21a5638c7f244737d8cab27dad348acb8d143023.png)
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
Proof with Mass Points
Let's First define some points' masses.
,
, and
By Mass Points:
The mass at A is
Multiplying them together,
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.
![[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]](http://latex.artofproblemsolving.com/8/7/e/87e7bd56f19d2bc7aa2f8aa9681e95b0d546ef76.png)