The angle condition implies the circumcircles of the three rectangles concur at a single point $P$. ! Then $\measuredangle C P B_{2}=\measuredangle C P A_{1}=90^{\circ}$, hence $P$ lies on $A_{1} B_{2}$ etc., so we're done. Remark. As one might guess from the two-sentence solution, the entire difficulty of the problem is getting the characterization of the concurrence point.

Leonard my dude.png
We first claim that the three circles $(BCC_1B_2),$ $(CAA_1C_2),$ and $(ABB_1A_2)$ share a common intersection.
Let the second intersection of $(BCC_1B_2)$ and $(CAA_1C_2)$ be $X$. Then
$$\begin{align*} \angle AXC &= 360^\circ - \angle BXA - \angle CXB \\ &= 360^\circ - (180^\circ - \angle AB_1B + 180^\circ - \angle BC_1C) \\& = 180^\circ - \angle CA_1A, \end{align*}$$
which implies that $AA_1C_2CX$ is cyclic as desired.
Now we show that $X$ is the intersection of $B_1C_2,$ $C_1A_2,$ and $A_1B_2.$ Note that $\angle C_1XB = \angle BXA_2 = 90^\circ,$ so $A_2, X, C_1$ are collinear. Similarly, $B_1, X, C_2$ and $A_1, X, B_2$ are collinear, so the three lines concur and we are done.
~Leonard_my_dude