It's well know $ABCD$ is a isosceles trapezium, so by symmetry $AD = BC$ and $AC = BD$. Notice that $AA_{1} = CA_{2}$ and $BB_{1} = CB_{2}$. Since $A_{1}B_{1}CD$ is cyclic, $\angle AA_{1}D = 180^{\circ} - \angle DA_{1}C = 180^{\circ} - \angle DB_{1}C = \angle BB_{1}D$ and as $\angle DAA_{1} = \angle DAC = \angle DBC = \angle DBB_{1}$ we have that $\triangle ADA_{1} \sim \triangle BDB_{1}$ where $\frac{CA_{2}}{CB_{2}} = \frac{AA_{1}}{BB_{1}} = \frac{AD}{BD} = \frac{BC}{AC}$. Then $CA_{2} \cdot CA = CB_{2} \cdot CB$, which by PoP at $C$ implies that $ABA_{2}B_{2}$ is cyclic $\blacksquare$