Clearly , the concurrency condition is equivalent to $ \frac { AK } { KB} = \frac { DL } { C L } $ and let $ AB $ and $ CD $ intersect in $ X $ and $ KM $ intersect $ NL $ in $ S $ . Then , the given condition imply that $ X KL S $ is cyclic and call this circle $ \gamma $ . Let $ O $ be spiral center of symmetry that sends $ A $ to $ D $ and $ B $ to $ C$ so it sends $ K $ to $L $ . So the point $O \in \gamma $ . Let $ T $ be the midpoint of $ KL $ and let $ MN $ intersect $ CD $ in $ G $ . There is a spiral symmetry that sends $ A $ to $ D $ , $ K $ to $ L $ , a spiral symmetry which sends $ A $ to $ K $ , $ D $ to $ L $ , a spiral symmetry that sends $ M $ to $ T $ and $ D $ to $ L $ and finally a spiral symmetry that sends $ D $ to $ M $ and $ L $ to $ T $ , all having the same center $ O $ . Consider the spiral symmetry $ H_{O} $ with center $ O $ which sends $ M $ to $ D $ and $ T $ to $ L $ . Since $ K $ must go $ K' $ in $ H_{O} $ such that $ \angle OKL = \angle OKT = \angle OK'L $ and so $ OKK' C $ form a cyclic quadrilateral and so $ K' \in \gamma $ . Let the line $ K'D $ intersect $ KM $ in $ S' $ . The spiral symmetry that sends $ K $ to $ K' $ and $ M $ to $ D $ has center $ O $ so the spiral symmetry that sends $ M $ to $ K $ and $ D $ to $ K' $ has also center $ O $ . So , the point $ S' $ belongs to the circle of $ K' K O $ and also the circle of $ D M O $ . So $ S' in $ circle $ \gamma $ and $ S' \in AD $ imply that $ S = S' $ . So , quadrilateral $ OSDM $ is cyclic . But the center of spiral symmetry that sends $ M $ to $ N $ and $ D $ to $ C $ must lie on the circle passing through $ G $ , $ D $ and $ M $ so also $ OGDM $ is cyclic and so $ SDMG $ is cyclic which proves the first equality of the problem and similarly we can show the second equality of the problem .

Let $G$ be a midpoint of $DB$ and $S$ point of intersection of $MK$, $DB$, $LN$. Then $\angle GMK=\angle MKA$ and $\angle GNS= \pi-\angle NLC=\pi-\angle MKA$, thus $MSNG$ is cyclic quadrilateral. Hence $\angle LNM= \pi-\angle MNS=\pi-\angle MGS= \angle MGD= \angle ABD$. By analogy we show that $\angle BDC=\angle KMN$.

Let $\rho$ be the spiral similarity which sends $D \rightarrow A, C \rightarrow B$. Then, let $A', B', M', N'$ be the images of $A, B, M, N,$ respectively, under this spiral similarity. Observe that we have from the concurrency that $L \rightarrow K$ (easy Menelaus), and so hence $\angle N'KB = \angle NLC = \angle AKM,$ i.e., $K \in MN'.$ Therefore, we have that $\angle KMN = \angle N'MN = \angle BDC$ by using Mean Geometry on $\triangle ABB', \triangle DCB.$ Analogously, Mean Geometry on $\triangle ABD, \triangle A'B'A,$ gives that $\angle M'N'M = \angle ABD \Rightarrow \angle LNM = \angle KN'M' = \angle M'N'M = \angle ABD.$ $\square$