Given is a convex $ABCD$, which is $ |\angle ABC| = |\angle ADC|= 135^\circ $. On the $AB, AD$ are also selected points $M, N$ such that $ |\angle MCD| = |\angle NCB| = 90^ \circ $. The circumcircles of the triangles $AMN$ and $ABD$ intersect for the second time at point $K \ne A$. Prove that lines $AK $ and $KC$ are perpendicular.
(Irán)
Define $B', D'$ to be $CN \cap AB, CM \cap AD$ respectively. Let $M_b, M_d$ be the midpoints of segments $BB', DD'$ respectively. Observe that $\triangle BCB' \sim \triangle DCD'$ since they're both $45-45-90$ triangles. Hence, as $\angle MCB = 90 - \angle BCD = \angle DCN$, we have that $\triangle BMC \sim \triangle DCN$ and $\frac{BM}{BB'} = \frac{DN}{DD'}.$ Since $K$ is the center of the spiral similarity sending $BM$ to $DN$, the previously discovered length ratio shows that it also sends $B'$ to $D'$, and so hence $K = (\triangle ABD) \cap (\triangle AB'D').$ As we also obviously have that $\frac{BM_b}{BB'} = \frac{DM_d}{DD'}$ since they're both $\frac12$, we've that $K = (\triangle ABD) \cap (\triangle AB'D') \cap (\triangle AM_bM_d).$ However, observe that $\angle AM_b = \angle AM_dC,$ which implies that $AC$ is the diameter of $(\triangle AM_bM_d)$, hence $\angle AKC = 90,$ as desired.
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