We want to investigate correspondences between triangles and orthocenters. Our first orthocenter is 'H' for triangle ABC. This point determines our next three triangles  AHC, AHB, and BHC.
Next, we construct a circumcenter by constructing perpendiculars through the midpoints of each segment.


Next, let's look at an arbitrary inscribed triangle  BHC. Whatever we show for it, we will be able to generalize to the rest of our shapes. So, let's construct its circumcenter.


This leads to an interesting observation:
Circumcenter of ABC is collinear with the Midpoint of CB and the Circumcenter of BHC 
This is because both of the circumcenters lie on the perpendicular to side CB which is in turn determined by the midpoint.
Not only that, but the middle point is the midpoint of the other two.


What we want is for Triangle CBI to be congruent to CHB. We will achieve this indirectly by looking at the similar triangles CKH and AJH (vertical angles and right angles).


Now we know that angle IAB and angle ICB are congruent because they subtend the same chord IB. By transitivity, since IAB = HAB which is congruent to HCK, we get HCK congruent to ICK. We also know that CKH is congruent to CKI (right angles), and CK is congruent to itself, so triangle CKH is congruent to CKI.
By using a similiar method, we can show that triangles IKB and HKB are congruent, so:

And no matter how hard I look, I still can't see why the two circumcenters are equidistant. Dangit.