Points Symmetric to the Orthocenter with Respect to the Sides of a Triangle

HA''	≈	1.31	HB''	≈	1.50	HC''	≈	6.52
A'A''	≈	1.31	B'B''	≈	1.50	C'C''	≈	6.52

Let ABC be a triangle. The points A', B', and C' symmetric to the orthocenter H with respect to the sides of ABC lie on the circumcircle. (A'', B'', and C'' are the feet of the perpendiculars from H to BC, CA, and AB, respectively.)