12.57*. In triangle $A B C$, the bisector $B E$ is drawn and a point $K$ is taken on side $B C$ such that $\angle A K B=2 \angle A E B$. Find the measure of angle $A K E$, if $\angle A E B=\alpha$.
12.57. Let $\angle ABC = 2x$, then the external angle $A$ of triangle $ABE$ is $\angle ABE + \angle AEB = x + \alpha$. Further, $\angle KAE - \angle BAK = (180^\circ - x - \alpha) - (180^\circ - 2x - 2\alpha) = x + \alpha$. Therefore, $AE$ is the bisector of the external angle $A$ of triangle $ABK$. Since $BE$ is the bisector of the internal angle $B$ of this triangle, $E$ is the center of its excircle, touching side $AK$. Hence, $\angle AKE = \angle AKC / 2 = 90^\circ - \alpha$.