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Let ABC be a triangle with orthocenter H and let

be the circumradius of ABC. Then:

AH²+BH²+CH²=12R²-(AB²+BC²+AC²)

See Theorem 204 in N. Altshiller-Court, College Geometry, 2nd ed., Mineola, NY: Dover, 2007 p. 102.

Contributed by: Jay Warendorff

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The Sum of the Squares of the Distances from the Vertices to the Orthocenter