Difference between revisions of "Menelaus' Theorem"
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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}=-\frac{PC}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{PC}= | + | <math>\frac{AR}{RB}\cdot\frac{QC}{QA}=-\frac{PC}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{PC}=1</math> |
== See also == | == See also == |
Revision as of 12:28, 26 July 2010
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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: