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

Let ABC be a triangle with orthocenter H and let be the circumradius of ABC. Then:
AH2+BH2+CH2=12R2-(AB2+BC2+AC2)

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