Bisection of Segments by the Sides of a Triangle
Let ABC be a triangle with orthocenter H, and let the feet of the altitudes be A', B' and C'. Let the extensions of the altitudes intersect the circumcircle a second time at A'', B'', and C''. Then HA' = A' A'', HB' = B' B'', and HC' = C' C''.
See Theorem 178 in N. Altshiller-Court, College Geometry, 2nd ed., Mineola, NY: Dover, 2007 p. 95.