Let $K_1, K_2, K_3, K_4$ be intersection points of line $A_1A_2$ and $B_1B_2, B_1B_2$ and $C_1C_2, C_1C_2$ and $D_1D_2, D_1D_2$ and $A_1A_2$ respectively. By angle chasing: $\angle BAD + \angle BCD = 180^{\circ} \Rightarrow \angle AA_1A_2 = \frac{180^{\circ} - \angle BAD}{2} = \frac{\angle BCD}{2} = \angle K_1A_1B$ Similarly, we get $\angle K_1B_2A_1 = \frac{\angle ADC}{2}$ So, $\angle B_2K_1A_1 = \angle K_2K_1K_4 = \frac{\angle BCD + \angle ADC}{2}$ Similarly, $\angle C_1K_3D_2 = \frac{\angle ABC +\angle BAD}{2}$ Then we get that $K_1K_2K_3K_4$ is cyclic. So we're done Note that points $A_1, B_2$ are on segment $AB$, $B_1, C_2$ are on segment $BC$, $C_1, D_2$ are on segment $CD$, and points $D_1, A_2$ are on segment $AD$. I think this should be given in the problem, anyways.