Let $\Gamma$ be the circle through $K,E,D$. Define points $X = BK \cap CD$ and $Y = BK \cap EM$.

From equal heights of parallelograms $BEMA$ and $ECDM$, we have $BY=YX$. Hence, by $SAS$ congruence we establish $\triangle BYE \equiv \triangle XYE$, and hence $BE = EX=MD$.

Thus, as $EX=MD$ and also $EM \parallel XD$ we have $EMDX$ as an isosceles trapezium and hence cyclic.

Furthermore, from alternate angles, $90^\circ = \angle ABY = \angle YXB \Longrightarrow \angle KXD = 90^\circ$.

Hence, as $\angle KED = \angle KXD = 90^\circ$, by Bowtie Theorem, $KEXD$ is cyclic.

Finally, we have that $EMDX$ and $KEXD$ share the same circle, so $MKEXD \Rightarrow MKED$ cyclic.

Let $\Gamma$ be the circle through $K,E,D$. Define points $X = BK \cap CD$ and $Y = BK \cap EM$.

From equal heights of parallelograms $BEMA$ and $ECDM$, we have $BY=YX$. Hence, by $SAS$ congruence we establish $\triangle BYE \equiv \triangle XYE$, and hence $BE = EX=MD$.

Thus, as $EX=MD$ and also $EM \parallel XD$ we have $EMDX$ as an isosceles trapezium and hence cyclic.

Furthermore, from alternate angles, $90^\circ = \angle ABY = \angle YXC \Longrightarrow \angle KXD = 90^\circ$.

Hence, as $\angle KED = \angle KXD = 90^\circ$, by Bowtie Theorem, $KEXD$ is cyclic.

Finally, we have that $EMDX$ and $KEXD$ share the same circle, so $MKEXD \Rightarrow MKED$ cyclic.