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

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!