Let $\odot(ABN)$ intersect $AD$ at $A_1$ and $BC$ at $B_1$; analogeously let $\odot(CDN$ intersect $BC$ at $C_1$ and $AD$ at $D_1$. Because of the cyclic quadrilaterals, we have $\sphericalangle{NA_1A}=\sphericalangle{NBA}=\sphericalangle{DCN}=\sphericalangle{DD_1N}$, implying that $\triangle{A_1ND_1}$ is isosceles with $NA_1=ND_1$. Analogeously, we get that $\triangle{B_1NC_1}$ is isosceles and therefore $NB_1=NC_1$. Last but not least, we see that due to $\sphericalangle{B_1A_1N}=\sphericalangle{CBN}=\sphericalangle{NAD}=\sphericalangle{NB_1A_1}$, triangle $\triangle{A_1B_1N}$ is isosceles with $NA_1=NB_1$. So altogether, we get $NA_1=NB_1=NC_1=ND_1$, proving the desired.