Proof. We have $\angle OB'C' = \angle OAC' = 90^\circ - \angle C$ and $\angle OB'A' = \angle OCA' = 90^\circ - \angle A$. So $\angle C'B'A' = \angle B$. Compute the other angles similarly. This way it is proved that the triangles are similar and $O$ is the orthocenter of triangle $A'B'C'$. Now, the altitudes of triangle $A'B'C'$ make the same angle with the corresponding sides of triangle $ABC$ (angles $\angle OB'A$ and so on). So, the corresponding sides of the triangles also make the same angle.

Lemma 2. Let $B'$, $C'$ be points on $AC$, $AB$ respectively such that $AB'OC'$ is cyclic. The circle $w_1$ with center $B'$ passing through $C$ intersects the circle $w_2$ with center $C'$ passing through $B$ at points namely $P$, $Q$ such that $P$ is on the circumcircle of $ABC$ and $Q$ is on $BC$. Moreover, $PQ$ and the altitude of $A$ intersect on the circumcircle of $ABC$.

Proof. Let $P$ be the second intersection of the circumcircles of triangles $ABC$ and $AB'C'$. We claim $P$ is on both $\omega_1$, $\omega_2$. We have
$$\begin{align*} \angle PBA &= \angle PCA = \frac{1}{2}\overarc{PA} = \frac{1}{2}\angle POA, \\ \angle PC'A &= \angle PB'A = \angle POA. \end{align*}$$
So, triangles $C'PB$ and $B'PC$ are isosceles and $P$ is on both $\omega_1$, $\omega_2$. Also, these triangles are similar. Let $\alpha = \frac{1}{2}\angle POA$.
Triangles $PBC$ and $PC'B'$ are similar (Consider the angles of $P$ and the ratio of the sides incident with $P$ in these triangles). The ratio of similarity is $2\cos\alpha$ and the corresponding sides make angles equal to $\alpha$. So, if $D$, $D'$ are the orthogonal projections of $P$ on $BC$ and $C'B'$ respectively, then $\frac{PD}{PD'} = 2\cos\alpha$ and $\angle D'PD = \alpha$. So triangles $D'PD$ and $C'PB$ are similar. So, $D'P = D'D$. Thus, if $PD'$ intersects $BC$ at $Q$, then $D'$ is the midpoint of the hypotenuse in the right angled triangle $PDQ$. So, $B'C'$ is the perpendicular bisector of $PQ$ and $Q$ is the second intersection of $\omega_1$, $\omega_2$.
Suppose $PD$ and $AH$ intersect the circumcircle of triangle $ABC$ for the second time at $E$ and $A_1$ respectively. We have
$$\angle QPE = \angle D'PD = \alpha = \frac{1}{2}\overarc{PA} = \frac{1}{2}\overarc{A_1E} = \angle A_1PE.$$
So, $PQ$ passes through $A_1$ and the lemma is proved. $\square$

According to lemma 2, the angle between $PQ$ and $AH$ is equal to the angle between $B'C'$ and $BC$. So, by lemma 1 the radical axes in the problem are obtained by rotating the altitudes of triangle $ABC$ with a fixed angle (but with different centers). The orientations of the rotations are the same, because all the equations in the proof remain valid by considering orientations. So, the triangle formed by the radical axes is similar to triangle $ABC$. Therefore, it suffices to prove that the ratios of distances of $H$ from the radical axes is the same as the ratios of distances of $H$ from sides of triangle $ABC$.

Let $F$, $A_2$ be the orthogonal projections of $H$ on $PQ$ and $BC$ as in the lemma. We have $\frac{HF}{HA_2} = 2\frac{HF}{HA_1} = 2\sin\alpha$ which is the same as the other ratios of distances of $H$ from the sides of the mentioned triangles. So the assertion is proved. $\square$