Let $H$ be the orthocentre of an acute triangle $ABC$. Consider the circumcenters of triangle $ABH,BCH,CAH$ .Prove that they are the vertices of a triangle that is congruent to $ABC$.
My Work:
Observe that $AH$ is perpendicular to $BC$ & $O_BO_C$. (Let $O_B$ be the circumcentre of $AHC$ and the others are defined same). Clearly,$AH\perp BC$ & $AH\perp O_BO_C$. Therefore $BC||O_BO_C$
Similarly $AC||O_AO_C$ & $AB||O_AO_B$
But, I failed to show $O_BO_C=BC$
You may also show that $AO_C=AO_B$
On
HINT:
Let $O$ be the center of the circle through $ABC$. Let $A'$ be the intersection of $AO$ with this circle. Check that $BA'CH$ is a parallelogram. Therefore $A'$ is the symmetric of $H$ with respect to the middle of $BC$. Now note that the circle $ABC$ is the circle $A'BC$. Now $HCB$ is the symmetric of $A'BC$ with respect to the middle of the segment $BC$. Therefore, $O_A$ is the symmetric of $O$ with respect to the middle of $BC$. With vectors, $\vec{OO_A}=\vec{OB}+\vec{OC}$. Similarly for $\vec{OO_A}$, $\vec{OO_A}$. Now it is easy to check that $\vec{O_AO_B}=-\vec{AB}$ and the others. In fact, the triangle $O_AO_BO_C$ is the symmetric of the triangle $ABC$ wr to the point $P$ such that $\vec{OP}=\frac{1}{2}(\vec{OA}+\vec{OB}+\vec{OC})$ ( the middle of the segment $OH$ as it turns out, see Euler line ).
$O_A O_B$ is the perpendicular bisector of $CH$, which is parallel to $AB$.
It follows that $ABC$ and $O_A O_B O_C$ are similar. Additionally, a dilation centered at $H$ with factor $2$ maps $O_A O_B O_C$ into the anticomplementary triangle of $ABC$, hence $O_A O_B O_C$ and $ABC$ are congruent.
Much more can be said: for instance, since $H$ is the circumcenter of $O_A O_B O_C$, the lines $AO_A,BO_B,CO_C$ are concurrent in the midpoint of $OH$, center of the nine point circle of $ABC$.