It is well-known that $EK$ is a symmedian of $\triangle AKC$. Let $EK$ and $AC$ meet at $P$. We have $\angle AKP=\angle CKN=\frac12(CM)$ so \[\angle PKN=\angle AKC-\angle AKP-\angle CKN=90^{\circ}+\frac12(BC)-\frac12(CM)-\frac12(CM)=90^{\circ}\] as desired.

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Let O be the midpoint of AB ONC=90 and NOK=90 OAE=90 Due to the similarity of triangles ANE and NOA we have NA.NA =NE.NO and AKCM is cyclic AN.AN=AN.NC=MN.NK we have MN.NK=NE.NO therefore EMOK is cyclic. Hence we have the result.