(Mathematics - Affine Geometry)

The theorem
For an arbitrary triangle ABC and an arbitrary point Q, let A' be the midpoint of the side opposite of A, B' the midpoint of the side opposite of B, C' the midpoint of the side opposite of C, a the line passing A' parallel to lAQ, b the line passing B' parallel to lBQ, c the line passing C' parallel to lCQ.
  1. The lines a, b, c are concurrent at a point P.
  2. The centroid G of triangle ABC lies on lPQ and 2(P - G) = G - Q.
Diagram
               P           C
               | \ a      /\     
              b|    \   /   \    
               |      /\     \   
               |    /     \   \  
               |  /          \ \ 
             B'./               \. A'
              /                  \     
            /                     \    
          /                        \   
        /                           \  
      /                              \ 
    /                C'               \
   A-----------------.-----------------B
     \                                 |
        \                              |
           \                           |
              \                        |
                 \                     |
                    \                  |
                       \               |
                          \            |
                             \         |
                                \      |
                                   \   |
                                      \|
                                       Q
(lines lCQ and c omitted to avoid clutter)

Preliminary observations
One of the preliminary results needed to prove this theorem is that a central dilatation is a collineation (a bijection that maps lines into lines). A tedious grinding through algebra shows this.

Another observation needed is the location of the centroid G of a triangle ABC.
G = (1/3)(A + B + C)    (see: centroid)
G = (1/3)A + (2/3)(1/2)(B + C)
This shows that G lies on the line that passes through A (see: equation for a line in Affine Geometry) and A' = (1/2)(B + C), and also that the distance from G to A is 1/2 the distance from G to A'.

Proof
The central dilatation δG,-1/2 maps triangle ABC into triangle A'B'C' due to the location of the centroid as mentioned. The central dilatation also maps lines lAQ, lBQ, and lCQ into a, b, and c since they are parallel. Hence δG,-1/2(Q) must lie on all three lines a, b, and c, which is the concurrent point P. Since δG,-1/2(Q) = P, it follows that (P - G) = (-1/2)(Q - G).
Q.E.D.

Result
This theorem shows that the centroid, the orthocenter, and the circumcenter of a triangle are collinear and lies on the Euler line when the line exists because each altitude is parallel to the perpendicular bisector of the same side of a triangle.

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