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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.
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:
The line through and is given by:
Which yields, after simplification,
$Z\cdotPR = -X\cdot(R-1)(P-1)+Y\cdotR(1-P)$ (Error compiling LaTeX. ! Undefined control sequence.)
Plugging in the coordinates for yields: