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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

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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.

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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!

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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.

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