Since the triangles $AA_1A_2, BB_1B_2, \Gamma\Gamma_1\Gamma_2$ and $\Delta\Delta_1\Delta_2$ are isosceles, using small letters for their equal angles we have the equalities:

Figure 4
$$\hat{A} + \hat{x} + \hat{x} = 180^\circ \Leftrightarrow \hat{x} = 90^\circ - \frac{\hat{A}}{2} \quad (1) \qquad \hat{B} + \hat{y} + \hat{y} = 180^\circ \Leftrightarrow \hat{y} = 90^\circ - \frac{\hat{B}}{2} \quad (2)$$
$$\hat{\Gamma} + \hat{z} + \hat{z} = 180^\circ \Leftrightarrow \hat{z} = 90^\circ - \frac{\hat{\Gamma}}{2} \quad (3) \qquad \hat{\Delta} + \hat{\omega} + \hat{\omega} = 180^\circ \Leftrightarrow \hat{\omega} = 90^\circ - \frac{\hat{\Delta}}{2} \quad (4).$$
Since the quadrilateral $AB\Gamma\Delta$ is cyclic, we have:
$$\hat{A} + \hat{\Gamma} = 180^\circ \Rightarrow \hat{x} + \hat{z} = 90^\circ \quad (5) \qquad \hat{B} + \hat{\Delta} = 180^\circ \Rightarrow \hat{y} + \hat{\omega} = 90^\circ \quad (6).$$
Let now the lines $A_1A_2, B_1B_2, \Gamma_1\Gamma_2$ form the quadrilateral $K\Lambda MN$. From the triangle $KA_1B_2$, we have: $\hat{K} + \hat{x} + \hat{y} = 180^\circ$, while from the triangle $M\Gamma_1\Delta_2$, we have: $\hat{M} + \hat{z} + \hat{\omega} = 180^\circ$. Summing up the last two relations and using (5) and (6) we obtain: $\hat{K} + \hat{M} = 180^\circ$.