# (BAMO $2013/3$) $ABH$, $BCH$ and $CAH$ is congruent to $ABC$.

Let $$H$$ be the orthocenter of an acute triangle $$ABC$$. Prove that the triangle formed by the circumcenters of triangles $$ABH$$, $$BCH$$ and $$CAH$$ is congruent to $$ABC$$.

I have already seen many answers on MSe But my doubt is diffferent,

In this solution (first one ) https://artofproblemsolving.com/community/c618937h1628954_problem_320_bamo_20133

To prove $$O_AB || O_BA$$, we can do some angle calculations: $$\angle O_ABC = 90 - A$$, and $$\angle C A O_B = 90 - B$$

how he got $$\angle O_ABC = 90 - A$$, and $$\angle C A O_B = 90 - B$$ ? I tried some angle chasing but did not able to get this ..thankyou

Extend $$BO_A$$ to $$D$$ such that BD is the diameter of the circle $$O_A$$.

Then, $$\angle BCD = 90^0$$.

Also, $$\angle 1 = \angle 2 = \angle 3$$

Then, purple marked ansgles are equal.

Because $$AO_B=CO_B$$, $$\measuredangle AO_BC=2\beta$$ and from here: $$\measuredangle CAO_B=90^{\circ}-\beta.$$ Another way.

Let $$A'$$, $$B'$$ and $$C'$$ be circumcenters of $$\Delta BHC$$, $$\Delta CHA$$ and $$\Delta AHB$$ respectively.

Thus, since $$B'HA'C$$ is a rhombus and $$HC=c\cot\gamma$$ in the standard notation, we obtain: $$A'B'=2\sqrt{B'H^2-\left(\frac{HC}{2}\right)^2}=2\sqrt{R^2-\frac{1}{4}c^2\cot^2\gamma}=$$ $$=2\sqrt{\frac{a^2b^2c^2}{16S^2}-\frac{1}{4}c^2\cdot\frac{\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}{\frac{4S^2}{a^2b^2}}}=2\sqrt{\frac{a^2b^2c^2}{16S^2}-\frac{c^2(a^2+b^2-c^2)^2}{64S^2}}=$$ $$=\frac{c}{4S}\cdot\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}=\frac{c}{4S}\cdot\sqrt{\sum_{cyc}(2a^2b^2-a^4)}=c.$$ Similarly $$A'C'=b$$ and $$B'C'=a$$ and we are done!

As it seems that you are in the EGMO forum, you should recall that:

The reflection of the orthocenter $$H$$ of $$\triangle ABC$$ over line $$BC$$ is the intersection of $$AH$$ and the circumcircle $$(ABC)$$.

Then let $$AH$$ and $$(ABC)$$ intersect again at $$D$$. By the lemma, the circumcenter of $$BCD$$ is $$O$$, and upon reflecting about $$BC$$, we obtain that $$O_A$$ is just the reflection of the circumcenter. Then $$\angle O_ABC=\angle OBC=90-\angle A$$, as needed.