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 = CP\cdot QA\cdot RB</math>. | Alternatively, when written with [[directed segment|directed segments]], the theorem becomes <math>BP\cdot CQ\cdot AR = CP\cdot QA\cdot RB</math>. | ||
+ | Also, the theorem works with all three points on the extension of their respective sides. | ||
== Proofs == | == Proofs == |
Latest revision as of 02:01, 12 June 2023
Menelaus' theorem deals with the collinearity of points on each of the three sides (extended when necessary) of a triangle. It is named after Menelaus of Alexandria.
Contents
[hide]Statement
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 . Also, the theorem works with all three points on the extension of their respective sides.
Proofs
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
QED
Proof with Mass points
First let's define some masses.
, , and
By Mass Points: The mass at A is Multiplying them together,
Converse
The converse of Menelaus' theorem is also true. If in the below diagram, then are collinear. The converse is useful in proving that three points are collinear.