Proof that the circumcenters of sub triangles forms a triangle congruent with the original triangle

Question

Let $\triangle ABC$ be a triangle with orthocenter H and let $O_A, O_B, O_C$ be the circumcenters of triangles $\triangle BCH, \triangle CAH, \triangle ABH$, respectively. Prove that the $\triangle O_AO_BO_C$ is congruent with $\triangle ABC$.

Can anyone help walk me through the proof?

Answer 1

Please insert a figure always, this shows the effort to understand the problem.

We consider thus the triangle $\Delta ABC$ with orthocenter $H$. Let $A', B', C'$ be the mid points of the segments $AH$, $BH$, $CH$. Then $O_A$, the circumcenter of the triangle $\Delta HBC$ is at the intersection of the side bisectors of the sides $HB$ and $HC$, so $O_A$ is the intersection of the perpendiculars in $B'$ and $C'$ on $HB$, respectively $HC$. So

• $O_A$ and $O_B$ are on the perpendicular in $C'$ on $CH$. So $O_AO_B$ is this side bisector
• Similarly, $O_BOC$ is the side bisector of $AH$, passing through $A'$, and
• similarly, $O_CO_A$ is the side bisector of $BH$, passing through $B'$.

The sides of $\Delta ABC$ and $O_AO_BO_C$ are thus respectively parallel, being in pairs perpendicular on the heights of $\Delta ABC$. So we have a similarity of the two triangles. Tho show their congruence (i.e. "equality") we need one more "metric relation". Well, let us get them all as follows. First of all $$\frac 12= \frac{B'C'}{BC}= \frac{C'A'}{CA}= \frac{A'B'}{AB}\ , $$ relations of mid segments in the triangles $\Delta HBC$, $\Delta HCA$, $\Delta HAB$.

Furthermore we have $B'C'\| O_BO_C$, and the similar relations $C'A'\|O_CO_A$, $A'B'\|O_AO_B$. How is then the triangle $\Delta A'B'C'$ placed w.r.t. $\Delta O_AO_BO_C$? Is it formed by mid segments? Yes, because for instance, using the parallelities: $$ \frac{O_AB'}{B'O_C} = \frac{O_AC'}{C'O_B} = \frac{O_CA'}{A'O_B} = \frac{O_CB'}{B'O_A} \ . $$ Thew two proportions at the beginning, and the end are reciprocal. So $C'$ is the mid point of $O_AO_B$. And the other relations. We get then the metric relation(s) $BC = 2B'C'=O_BO_C$ (and the other two).

$\square$

Bonus: The height in $O_A$ in $\Delta O_AO_BO_C$ is also perpendicular on $B'C'$, so it is the side bisector of $B'C'$. And so it passes through the circumcenter of $\Delta ABC$. Same for the other vertices in $\Delta O_AO_BO_C$, so its orthocenter is the circumcenter of $\Delta ABC$. And conversely, as it is seen in the picture.