Consider the following setup:
1. Construct the orthocenter H of triangle ABC.
2. Construct the circumcenters A',B', and C' of triangles BHC, AHC, and AHB, respectively. Then draw circles ABC, BHC, AHC, and AHB.
Proposition: Circles ABC, BHC, AHC, and AHB are congruent.
Proof: Since the constructions for circles BHC, AHC, and AHB were identical, it will suffice to prove circle ABC and circle AHC are congruent. We draw the diagram again, showing the complete altitudes of triangle ABC (note: O is the center of circle ABC).
Our first goal is to show that angle AB'C is congruent to angle AOC. Suppose the measure of angle AB'C is 2x. Then angle AHC measures 180-x and thus angle PHQ measures 180-x. Since triangle PHB and triangle QHB are right triangles, the sum of angles PHQ and ABC must be 180. Thus angle ABC measures x and angle AOC measures 2x.
Triangles AOC and AB'C are isosceles triangles with congruent vertex angles and bases, therefore they are congruent. Now, the radii of our two circles are congruent so the circles are congruent.
Therefore, Circles ABC, BHC, AHC, and AHB are congruent.