The problem statement follows:
Let $H$ be the orthocenter of an acute angle triangle $ABC$. Consider the circumcenters of triangles $ABH$, $ BCH$, and $CAH$. Prove that they are the vertices of a triangle that is congruent to $ABC$.
So I first showed that the orthocenter of $ABC$ is the circumcenter of the second triangle, $A'B'C'$ and the circumcenter of $ABC$ is the orthocenter of $A'B'C'$. Next if we take a homothety $h$ at $N_9$ of $ABC$ with scale factor $-1$, this will send $H$ to $O$ and vice versa and we'll get a congruent triangle.
But my question is, how do I prove that the triangle formed by taking the homothety is the very triangle in the question, namely $A'B'C'$?
Best Answer
I believe you know the fact that circle $BCH$ is a reflection of circle $ABC$ acros $BC$. Similary is true for the other two circles.
So $$AC' = AO = CO = CA'$$
and by easy angle chase you can see $AC'||CA'$ so $A'CAC'$ is paralelogram, so $A'C' = AC$ and we are done.