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.
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
Proof with Mass points
First let's define some masses.
, , and
By Mass Points: The mass at A is Multiplying them together,
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.