Beautiful! Define $A_M$, $B_M$, and $C_M$ as the midpoints of $HA'$, $HB'$, and $HC'$ respectively, and denote by $H''$ the orthocenter of $\triangle A_MB_MC_M$. It suffices to prove that $H''$ is the circumcenter of the triangle. It is a well known fact that these points lie on the sides of the triangle. Now for the key claim: $\textbf{Claim.}$ The circle $(A_MH''B_M)$ passes through the vertex of the triangle opposite $l_c$. $\textit{Proof.}$ It is enough to prove that the angle between $l_a$ and $l_b$ is the same as $A'HB$. Define $A_1$ and $A_2$ as the feet of the altitudes from $A'$ to $BC$ and $AC$ respectively, and define $B_1$ and $B_2$ the same but with $B$. It is enough to show that $\measuredangle B'CA = \measuredangle A_1NB_1$. Indeed, \[ \measuredangle A_1NB_1 = \measuredangle B_2B_1C + \measuredangle CA_1A_2 = \measuredangle B_2B'C + \measuredangle CA'A_2 = 180^{\circ} - \measuredangle B'CA - \measuredangle ACA' = \measuredangle B'CA. \]The analogous proof is easily done but is way too tedious so lol. Now observe that by reflecting the orthocenter we obtain that the circles $(A_MH''B_M)$, $(B_MH''C_M)$, and $(C_MH''A_M)$ are congruent. If we let $l_b$ and $l_c$ intersect at $P_a$, and likewise for $P_b$ and $P_c$, it is not hard to see that the arcs that $\angle H''P_bA_M$ and $\angle H''P_cA_M$ subtend are the same, hence those angles are equal, or $H''P_a=H''P_b$. Repeat for the other ones, so we get $H''=O$. Apologies for the not so good point names as it is 1 AM right now (oof)

My solution during the test: Let $A_0 = l_b \cap l_c$, $B_0 = l_c \cap l_a$, $C_0 = l_c \cap l_a$, $A_1 = l_a \cap A'H$, $B_1 = l_b \cap B'H$, $C_1 = l_c \cap C'H$. $Lemma$ $1:$ $A_1$ is midpoint of $A'H$ (similar for $B$ and $C$). $Proof:$ The reflection of $A'$ by $AB$ is $x = a+b-ab\overline{a'}$ and by $AC$ is $y = a+c-ac\overline{a'}$. Also, $h = a+b+c$. So, by homothety of center $A'$, we just have to show that $x, y, h$ are collinear $\iff$ $$\frac{h-x}{h-y} = \frac{\overline{h}-\overline{x}}{\overline{h}-\overline{y}} \iff \frac{c+ab\overline{a'}}{b+ac\overline{a'}}=\frac{\overline{c}+\overline{ab}a'}{\overline{b}+\overline{ac}a'} \iff (c+ab\frac{1}{a'})(\frac{1}{b}+a'\frac{1}{ac})=(b+ac\frac{1}{a'})(\frac{1}{c}+a'\frac{1}{ab}) \iff 0 = 0. _\blacksquare$$ So, there is a homothety $\Phi$, with center at $H$ and ratio $2$, that sends $A_1$ to $A'$, $B_1$ to $B'$ and $C_1$ to $C'$. Let $O'$ be the midpoint of $HH'$, so, the homothety sends $O'$ to $H' \therefore$ $O'$ is the orthocenter of $A_1B_1C_1$. Then, we just have to prove that $O'$ is also the circumcenter of $A_0B_0C_0$. $Lemma$ $2:$ $A_1B_1C_1 \sim A_0B_0C_0$. $Proof:$ It suffices to show that $\angle C_1 = \angle C_0$, since the other angles are analogous. But, by the homothety $\Phi$, we have that $\angle C_1 = \angle C'$, so we just have to prove that $\angle C_0 = \angle C'$. Let $K=l_b \cap AC$, $L=l_b\cap AB$, $M=l_a\cap BC$ and $N=l_a\cap AB$. By Simson Line, $B'K \perp AC$ and $A'M \perp BC$. So, $\angle C_0 = \angle ACB + \angle C_0KC + \angle CMC_0 = \angle C + \angle C_0KC + \angle CMC_0$. Note that $\angle CMC_0 = \angle BMN = \angle BA'N = 90^\circ - \angle NBA' = 90 ^\circ -(180^\circ - \angle A'BA) = \angle A'BC + \angle B - 90 ^\circ$. Anagously, $\angle C_0KC = \angle CAB' + \angle A - 90 ^\circ$. And, finally, $\angle C_0 = \angle C + \angle A + \angle B - 180^\circ - \angle A'BC + \angle CAB' = \frac{\widehat{B'C}}{2}+\frac{\widehat{CA'}}{2}+\frac{\widehat{B'A'}}{2} = \angle C'. _\blacksquare$ With this in hands, we have that $\angle{C_1O'B_1} = 180^\circ - \angle C_1A_1B_1$ (orthocenter) $= 180^\circ - \angle B_1A_0C_1$ (Lemma 2) $\therefore A_0C_1O'B_1$ is ciclic, and, analogously, $B_0A_1O'C_1$ is ciclic. Then, $\angle O'A_0C_1 = \angle O'B_1C_1$ and $\angle C_1B_0O' = \angle C_1A_1O'$, but, as $O$ is the orthocenter of $A_1B_1C_1$, we have that $\angle{O'B_1C_1} = \angle{C_1A_1O'} \therefore$ $\angle{O'A_0C_1} = \angle{C_1B_0O'} \implies A_0B_0C'$ is isosceles and $A_0O'=B_0O'$. Similarly, we get that $A_0O'=B_0O' = C_0O' \implies O'$ is the circumcenter of $A_0B_0C_0$ and midpoint of $HH'. _\blacksquare$ PS.: For other configuration of points $A', B', C'$ the angle chasing is similar.

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Let $XYZ$ be the relevant triangle, and let $M_D, M_E, M_F$ be the midpoints of $\overline{HD}, \overline{HE}, \overline{HF}$. By Simson line bisection we just need to show that the orthocenter $R$ of $M_DM_EM_F$ is the circumcenter of $XYZ$. Claim. $X, M_E, R, M_F$ are concyclic. Proof. Boils down to $\angle YXZ = \angle FDE$ by the given orthocenter. The Simson lines are parallel to $\overline{BE'}$ and $\overline{BF'}$ where $E', F'$ are the reflections of $E, F$ about $\overline{AC}$. But $\angle E'BF' = \angle FDE$ clearly so the result follows. $\blacksquare$ Similarly, the two symmetric quadrilaterals are also cyclic. Then $\measuredangle M_DZR = \measuredangle M_DM_ER = \measuredangle M_DM_FR = \measuredangle M_DYR$, hence $RY = RZ$ and the result follows symmetrically.

Replace $\triangle A'B'C'$ with $\triangle XYZ$. Perform a homothety at $H$ with ratio $2$, so that we wish to show that the orthocenter of $\triangle XYZ$ is the circumcenter of the triangle formed by the three new lines. Use complex numbers. The image of $\ell_a$ is the line parallel to the Simson line of $X$ through $X$, Since the Simson line rotates at half the rate of the point on the circle, the second intersection of this image line with the circumcircle is $-\frac{abc}{x^2}$. Thus, using the intersection formula, the intersection of chords $x,-\frac{abc}{x^2}$ and $y,-\frac{abc}{y^2}$ is $$x+y+\frac{abc}{xy}.$$Subtracting $x+y+z$ from this this yields $$-z+\frac{abc}{xy}$$which multiplying in $xy$ which has magnitude $1$ is equal in magnitude to $-xyz+abc$, which is symmetric so all symmetric variants have the same magnitude. Thus, each of these intersections is the same distance from $x+y+z$, done.