Solution

One solution is based on a property of the orthocenter: its reflection in a side of the triangle falls on the circumcenter. In other words, the segment of an altitude extended between the orthocenter and the second intersection with the circumcircle is bisected by the corresponding side of the triangle.

Let ΔABC be given. H is the orthocenter and H_a the foot of the altitude from A (H_b and H_c are defined likewise.) Let AH_a cross the circumcircle the second time in, say, P. Then HH_a = PH_a, giving ΔHBC = ΔPBC. If two triangles are equal. so are their circumcircles. But the circumcircle of ΔPBC coincides with that of ΔABC thus proving the statement.

Douglas Rogers reports that the problem in a little different guise but with a superbly different solution has been posed in the Mathematical Repository (New Series) published by Thomas Leybourn (Question 377). Two solutions appeared in volume IV (1819), 91-92. Of these the second solution by the proposer (Mr. Cunliffe) was essentially as above. But the first one, by a mysterious Lady, was essentially different. The applet illustrates that solution. But first the problem as it appeared in the Mathematical Repository:

The answer is of course that all three are equal and, moreover, they are equal to the circumdiameter of the given plane triangle.

Let HA', HB', HC' be diameters of the three circles. Since ∠HBA' is subtended by a diameter in circle BCH, ∠HBA' = 90°. Similarly ∠HBC' = 90°. It follows that A'BC' is a straight line. Similarly, C'AB' and B'CA' are straight.

Further

Hence, ∠CB'H = ∠CA'H, making ΔA'B'H isosceles and HA' = HB'. Similarly we get, say, HB' = HC'.