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

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?

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.