[ Four points lying on one circle]

Diagonals $A C$ and $C E$ of a regular hexagon $A B C D E F$ are divided by points $M$ and $N$ such that $A M: A C=C N: C E=\lambda$. Find $\lambda$, given that points $B, M$ and $N$ lie on the same line.
Since $E D=C B, E N=C M$ and $\angle D E C=\angle B C A=30^{\circ}$ (fig.), then $\triangle E D N=\triangle C B M$. Let $\angle M B C=\angle N D E=\alpha, \angle$ $B M C=\angle E N D=\beta$. Clearly, $\angle D N C=180^{\circ}-\beta$. Considering triangle $B N C$, we get $\angle B N C=90^{\circ}-\alpha$. Since $\alpha+\beta=180^{\circ}-30^{\circ}=150^{\circ}$, then $\angle D N B=\angle D N C+\angle C N B=\left(180^{\circ}-\beta\right)+\left(90^{\circ}-\alpha\right)=270^{\circ}-(\alpha+\beta)=$ $120^{\circ}$. Therefore, points $B, O, N$ and $D(O-$ center of the hexagon) lie on the same circle. Since $C O=C B=$ $C D$, i.e., $C$ is the center of this circle, thus, $\lambda=C N: C E=C B: C A=1: \sqrt{3}$.

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