Suppose $X$ is the second intersection of circles $(ABC)$ and $(AB_1C_1)$. Note that $\ell_A$ is the perpendicular bisector of segment $XA$, then, it passes through the circumcenter $O$ of triangle $ABC$. We conclude that $\ell_A,\ell_B$ and $\ell_C$ concur at $O$.

Let $O$ be the circumcenter of $ABC$ and $P$ be the midpoint of $CH$, it suffices to show $OP \perp CC_2$. Solution 1 We shall use that the arbitrary lines $AB$ and $CD$ are perpendicular iff $AC^2 - AD^2 = BC^2 - BD^2$. Hence we wish to prove $OC_2^2 - OC^2 = PC_2^2 - PC^2$. By Stewart's theorem in $AOB$ (or power of a point) we get $OC_2^2 = R^2 + C_2A \cdot C_2B = OC^2 + C_2A \cdot C_2B$, similarly from $PA_1B_1$ we get $PC_2^2 = PA_1^2 + C_2B_1 \cdot C_2A_1 = PC^2 + C_2B_1 \cdot C_2A_1$. On the other hand, the cyclic $ABA_1B_1$ gives $C_2A \cdot C_2B = C_2A_1 \cdot C_2B_1$ and the result follows. Solution 2 Let $T$ be the diametrically opposite point of $C$ with respet to the circumcircle $k$ of $ABC$. Since $OP$ is a midsegment in $CHT$, it suffices to have $HT \perp CC_2$. If $HT\cap k = X$, then $\angle CXH = \angle CXT = 90^{\circ}$ and so it suffices to justify that $AB$, $A_1B_1$ and $CX$ are concurrent. But these actually are the radical axis of the circumcircles of $ABC$, $ABA_1B_1$ and $CA_1HB_1X$, so we are done. (Alternatively, similarly to what @Diegoo_1108 did, we may avoid constructing $T$ and instead define $X$ as the second intersection point of the circumcircles of $ABC$ and $A_1B_1C$. After arguing that $CX$, $AB$, $A_1B_1$ are concurrent radical axis, so that $C$, $X$, $C_2$ are collinear, it is enough to note that the line through centers, which is $OP$, has to be perpendicular to the radical axis, which is $CXC_2$.)