Difference between revisions of "Menelaus' Theorem"
Pi3point14 (talk | contribs) (→Proof Using Barycentric coordinates) |
Pi3point14 (talk | contribs) (→Proof Using Barycentric coordinates) |
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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\ | + | 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 | QED |
Revision as of 21:03, 28 August 2015
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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
where all segments in the formula are directed segments.
Proof
Draw a line parallel to through to intersect at :
Multiplying the two equalities together to eliminate the factor, we get:
Proof Using 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