I started by constructing the orthocenter, H, of any triangle ABC
Then constructed the circumcenters for triangles ABC, HBC, HAB, and HAC…
A cleaned up versions with just the circumcenters…
Here is a link to the GSP file that may come in handy.
Now let’s explore what happens when the orthocenter is the same as one of the triangle’s vertices…
Triangle HAB is now the same as Triangle CAB and therefore their circumcircles are congruent; also if the orthocenter lies on one of the vertices of the triangle, then it must form a right angle.
We see this is still true when H is now located at the triangle vertex B…
Now if we go back to the circumcircles and connect the centers with the vertices we get the shape below:
The radii of the circumcircles form isosceles triangles.
Also the radii of the circumcircles are equidistant from the orthocenter, as they all lie on the black circle centered at H:
Because the radii are all equal, the circles all have the same area.
I have also constructed the nine point circle for Triangle ABC:
After constructing the nine point circle for Triangle HAB, I realized it was the same nine point circle as Triangle ABC’s:
This makes sense because we are using the same perpendicular bisectors for each of the sides to find the circumcenters.
Here is the GSP file for the nine point circle that might come in handy.