the three medians of a triangle are concurrent; and the points
at which the medians intersect is one third of the way along each median measured
*towards* the vertex.

Synthetic proof: Consider the triangle ABC and let the midpoint of AB be F,
and that of AC be E; then BE and FC intersect at G, say. It's enough to show
that the line segment from A to the midpoint of BC -- call it D -- passes through G.

Construct a circle with center G; let H be the point diametrically opposite to A (hence
AG = GH).

Consider triangle ABH. Since F and G are midpoints, FG is parallel to BH
from the Midpoint Theorem; similarly GE is parallel to HC. Therefore,
BGCH is a parallelogram; and the diagonals BC and GH bisect each other, that is,
AH bisects BC at D. Finally, we see that the third median AD also passes through G, i.e.
the three ...

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