Difference between revisions of "Menelaus' Theorem"
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Alternatively, when written with [[directed segment|directed segments]], the theorem becomes <math>BP\cdot CQ\cdot AR =
Alternatively, when written with [[directed segment|directed segments]], the theorem becomes <math>BP\cdot CQ\cdot AR = PC\cdot QA\cdot RB</math>.
== Proofs ==
== Proofs ==
Revision as of 19:34, 14 December 2019
If line intersecting on , where is on , is on the extension of , and on the intersection of and , then
Alternatively, when written with directed segments, the theorem becomes .
Proof with Similar Triangles
Draw a line parallel to through to intersect at :
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
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.