\(ABC\) any triangle

Any \(BD\) and \(CE\) drawn such that \(D\) lies on \(AC\) and \(E\) lies on \(AB\)

\(DE\) drawn

\(BD\) and \(CE\) intersect at \(I\)

\(AO=OI\)

\(EN=ND\)

\(BM=MC\)

Prove that \(M-N-O\) are collinear

\(ABC\) any triangle

Any \(BD\) and \(CE\) drawn such that \(D\) lies on \(AC\) and \(E\) lies on \(AB\)

\(DE\) drawn

\(BD\) and \(CE\) intersect at \(I\)

\(AO=OI\)

\(EN=ND\)

\(BM=MC\)

Prove that \(M-N-O\) are collinear

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TopNewestWe will use normalized barycentric coordinates \(A = (1,0,0)\), \(B = (0,1,0)\), \(C = (0,0,1)\). Then \(D = (p,0,1 - p)\), \(E = (q,1 - q,0)\). Now \(N\) is a midpoint of \(DE\), hence \[N = \left(\frac{p + q}{2}, \frac{1 - q}{2},\frac{1 - p}{2}\right)\] Equation of line \(BD\) is \[pz = (1 - p)x\] and equation of line \(CE\) is \[(1 - q)x = qy\] Intersecting these lines we get \[I = \left(\frac{pq}{p + q - pq}, \frac{p(1 - q)}{p + q - pq}, \frac{q(1 - p)}{p + q - pq}\right)\] Now \(O\) is midpoint of \(AI\), hence \[O = \left(\frac{p + q}{2(p + q - pq)}, \frac{p(1 - q)}{2(p + q - pq)}, \frac{q(1 - p)}{2(p + q - pq)}\right)\] And \(M\) is midpoint of \(BC\), so trivially \[M = \left(0, \frac{1}{2}, \frac{1}{2}\right)\] Now note that \[\begin{vmatrix} \frac{p + q}{2}& \frac{1 - q}{2}&\frac{1 - p}{2} \\ \frac{p + q}{2(p + q - pq)}& \frac{p(1 - q)}{2(p + q - pq)}& \frac{q(1 - p)}{2(p + q - pq)} \\ 0& 1/2& 1/2 \end{vmatrix} = 0\] Hence \(M,N,O\) are collinear. – Jan J. · 3 years, 11 months ago

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– Sotiri Komissopoulos · 3 years, 11 months ago

I'm not sure I understand your last step. Are you taking the determinant of a matrix? I'm assuming that's what you're doing. How does that tell you that the three points are co-linear?Log in to reply

Note that once you found the coordinates of the points, you could go about finding the line that these three points are on. However, unless the ratios are immediate, this often requires a bunch more algebra. Showing that the area is 0 could be more straightforward.

Barycentric coordinates merely make the algebra much easier / nicer. A coordinate geometry approach would work similarly, though with much uglier values. – Calvin Lin Staff · 3 years, 11 months ago

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– Jan J. · 3 years, 11 months ago

See here http://zacharyabel.com/papers/Barycentric_A07.pdf page 5.Log in to reply

This is the Newton-Gauss line of a complete quadrilateral. – Jon Haussmann · 3 years, 11 months ago

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Why did every comment gets vote down? – Zi Song Yeoh · 3 years, 8 months ago

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