I like this problem Let $\omega = (BA_1C_1)$. Recall that $\overline{B_0A_1}$, $\overline{B_0C_1}$, and the line through $B$ parallel to $\overline{AC}$ are tangent to $\omega$. Therefore, $\omega$ is actually the incircle of $\triangle B_0A'C'$. [asy][asy] size(8cm); defaultpen(fontsize(9pt)); pair B = dir(110), A = dir(180+30), C = dir(-30), B0 = (A+C)/2, H = A+B+C, A1 = foot(A, B, C), C1 = foot(C, A, B), Ap = extension(A1, B0, B, rotate(90, B)*H), Cp = extension(C1, B0, B, rotate(90, B)*H), X = extension(A, Ap, C, Cp); draw(A--B--C--cycle, red+linewidth(0.9)); draw(Ap--Cp--B0--cycle, blue+linewidth(0.9)); draw(C--Cp^^A--Ap, gray); draw(A1--C1, heavygreen); draw(A--A1^^C--C1, gray); draw(circumcircle(A1, C1, B), fuchsia); string[] names = {"$A$", "$B$", "$C$", "$B_0$", "$H$", "$A_1$", "$C_1$", "$A'$", "$C'$", "$X$"}; pair[] points = {A, B, C, B0, H, A1, C1, Ap, Cp, X}; pair[] ll = {A, B, C, B0, N, dir(10), C1, Ap, Cp, N}; for (int i=0; i<names.length; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy] Let $X = \overline{AA'} \cap \overline{CC'}$. Note that $A_1, X, C_1$ are collinear by Pappus Theorem on $\overline{C'BA'}$ and $\overline{AB_0C}$. Moreover, since a homothety at $X$ sends $AC$ to $A'C'$, it also sends $B_0$ to the midpoint $M$ of $A'C'$; in particular, $B_0, X, M$ are collinear. Now it is well-known that $X$ must lie on the perpendicular line to $\overline{A'C'}$ through $B$.

Clearly $\omega = (BA_1C_1)$ is the incircle of $B_0A'C'$. Let $O$ be the center of $\omega$. Claim: $AO \perp CC'$ Proof: We use perpendicularity Lemma:\begin{align*}OC^2 - OC'^2 &= (OC^2 - R^2) - \ell^2 - h^2 = CH \cdot CC_1 - \ell^2 - h^2\\AC^2 - AC'^2 &= AC^2 - (AC_1 + \ell)^2 - h^2 = CC_1^2 - 2AC_1\ell - \ell^2 - h^2\end{align*}where $\ell = \tfrac12BC_1$ and $h = \delta(C', AB)$. Indeed, check\[CC_1^2 - 2\ell AC_1 = CC_1^2 - AC_1\cdot BC_1 = CC_1^2 - HC_1\cdot CC_1 = CH \cdot CC_1\]as desired. $\square$ Similarly, $CO \perp AA'$. So $C \in \text{polar } A' \iff A' \in \text{polar } C$, and since $OC \perp AA'$, it follows that $AA'$ is the polar of $C$. Similarly, $CC'$ is the polar of $A$, hence $X$ lies on both so $AC$ is the polar of $X$. Thus, $X \in BH$. $\blacksquare$

since $B_0C_1,B_0A_1$ are tangent to $(BA_1HC_1)$ let $R=BH \cap A_1C_1$ note that $RA$ is the polar of $C$ with respect to $(BA_1HC_1)$ and since $C \in polar(A')$ we have $A-R-A'$ are collinear , so are $C-R-C'$

Just wondering why is everyone overkilling this problem with polar?

another way to solve this is by considering $AA' \cap AH=X$ then we know that $a=\frac{AX}{XH}=\frac{S_{ABA'}}{S_{AHA'}}$ then by computing them(calculating $S_{ABA'}$ is easy and $S_{AHA'}=S_{AA_1A'}-S_{AHA'}$ and the rest is easy) we get that $a=tan C\times tan A$ which is symmetric with respect to $A$ and $C$ which means $AA' \cap AH=CC' \cap AH$