Let $A_1B_1 \cap A_2B_2 = X$. By homotheties we may see that $X$ is the top point of $\Gamma$. By Ceva it suffices to prove that $CX$ is the median of $CA_1A_2$, which is readily equivalent to $X$ being on the radical axis of $\Gamma_1, \Gamma_2$. But since $AX$ is a diameter of $\Gamma$, we have that an inversion from $X$ sends $A_i$ to $B_i$ for $i \in \{1, 2\}$, so thus $A_1A_2B_1B_2$ is cyclic and radical axis on $(A_1A_2B_1B_2), \Gamma_1, \Gamma_2$ finishes.

Here is an alternative solution that I smh did not notice to use Ceva. Define $A'$, $X_1$, $X_2$ as the point diametrically of $A$ in $\Gamma$, $A_1$ in $\Gamma_1$, $A_2$ in $\Gamma_2$, respectively. By homothety, $A_1, B_1, A'$ are collinear and $A_2, B_2, A'$ are collinear. Moreover $\triangle A'B_2A \sim \triangle A'AA_2$ from two right angles. Therefore, $A'B_2 \cdot A'A_2 = A'A^2 = A'B_1 \cdot A'A_1$, meaning $A_1, B_1, A_2, B_2$ are concyclic. Define $D_1' = A_1C \cap \overline{AB_1X_1}$ and $D_2' = A_2C \cap \overline{AB_2X_2}$. We know that $\measuredangle D_1'B_1D_2' = \measuredangle D_1'B_2D_2' = \measuredangle D_1'CD_2' = 90^{\circ}$. So the five points are concyclic. Then, $\measuredangle CB_2D_1' = \measuredangle CB_1D_1' = \frac 12 \widehat{CX_1} = \frac 12 \widehat{CA_2} = \measuredangle AB_2A_2$, meaning $D_1', B_2, A_2$ are collinear. So $D_1' = D_1$. Now apply Reim wrt $(A_1A_2B_1B_2)$ and $(B_1B_2D_1D_2C)$ on lines $A_1D_2$ and $A_2D_1$ yields the result.