Invert at $M$ with radius $1$. Let $X'$ denote the inverse of $X$. It suffices to show that $K',C',D'$ are collinear. For this, by Menelaus, it suffices to show that $\frac{K'B'}{K'A'} \cdot \frac{A'C'}{C'N'} \cdot \frac{N'D'}{D'B'} = 1$. By inversion distance formula, we have $K'A' = \frac{KA}{MA\cdot MK}$ and $K'B' = \frac{KB}{MK\cdot MB}$ so we have $\frac{K'B'}{K'A'} = \frac{MA}{MB}$. By a similar argument, we get $\frac{C'A'}{C'N'} = \frac{MN}{MA}$ and $\frac{N'D'}{D'B'} = \frac{MB}{MN}$. Thus, by Menelaus we have $K', C', D'$ are collinear and inverting back, we get $K,M,C,D$ are concyclic and since $\angle CMD = 90$, we have $\angle CKD = 90$ and $CL=KL=LD$. $\square$