Let $E,F,G$ be the intersections of a circle, with center in $L$ and radius $LB$, with $AB$, $AC$ (between A and C) and $BC$, respectively. Since $\angle EAK=\angle FAK=\alpha$ and $AK$ passes through the center of $(BEF)$ we have $\angle EBK=\angle FBK=30-\alpha\rightarrow \angle FBG=\alpha\rightarrow FLG=2\alpha$ (1). It is easy to see that $LKG$ is an equilateral triangle, so $\angle AEK=\angle KGB=60$, and since $F$ is the reflection of $E$ w.r.t. the bisector $AK$ we have that $\angle AFK=60\rightarrow KFCL$ is cyclical. So $\angle FKC=\angle FLC= (1)=2\alpha\rightarrow AMKF$ is cyclic (<MAF=2alpha), so $\angle EMK=60$ therefore $MKE$ is an equilateral triangle. We see that $KG$ is a bisector of $\angle FKC=2\alpha\rightarrow \angle BKF=\angle BKM=90+\alpha$ and if we consider that $MK=KF\rightarrow MBFN$ is a kite, therefore $\angle KMN=\angle KFN=60\rightarrow \angle AMN=60$.

Attachments: