What is M?

just added $\{C, M\} =C_3 \cap C_4$ at the end, thanks for noting the missing info

$Lemma$ Consider a triangle $ABC$ with $\angle C = 90º$. Let $\omega$ and $\lambda$ be the circles with diameter $AC$ and $BC$, take $M = \omega \cap \lambda, M \ne C.$ Thus, $M$ is the foot of the height from $C$ to $AB$. $Proof:$ Let $M'$ be the foot of the height from $C$ to $AB$, as $\angle CM'A = 90º \implies \omega$ is the circumcircle of $\triangle CM'A$ too, $\angle CM'B = 90º \implies \lambda$ is the circumcircle of $\triangle CM'B.$ So $CM'$ is the radical axis of $\omega$ and $\lambda$, hence $M \equiv M'. \blacksquare$ Applying the $Lemma$ with triangles $ABC, ADE, ACE, BCD$ we get that: $K, L, M, N$ are the feet of the heights from $C$ to $AB, BD, DE, EA$, respectively. So, $CK \perp AB$, $CL \perp BD$, $CM \perp DE$ and $CN \perp AE.$ Take $\alpha = \angle CBD$, $\beta = \angle CAE.$ Note that: $\angle NCA = 90º-\beta \implies \angle NCE = \beta.$ Too we have that: $\angle BCL = 90º-\alpha \implies DCL = \alpha.$ Besides that, the following quadrilaterals are cyclic, because of the perpendicularism said over: $CMLD, CMNE, CLKB$ and $CNKA$, thus: $\angle DML = \angle DCL = \alpha, \angle EMN = \angle ECN = \beta \implies \angle NML = 180º - \angle DCL + \angle ECN = 180º-(\alpha + \beta)$ now calculating $\angle NKL = \angle CKL + \angle CKN = \angle CBL + \angle CAN = \alpha + \beta.$ Therefore, $\angle NKL + \angle NML = 180º-(\alpha+\beta)+(\alpha+\beta) = 180º,$ so $K, L, M, N$ are concyclic because quadrilateral $NKLM$ is cyclic $.\blacksquare$