Answer. $\frac{A M}{A B}=\frac{3}{2 \sqrt{5}}$. Solution. Let $\angle A B M=\alpha$. Then $\angle B M C=2 \alpha, \angle B M A=180^{\circ}-2 \alpha$, $\angle B A M=180^{\circ}-\angle A B M-\angle A M B=\alpha=\angle A B M$, from which $B M=A M=M C$. It follows that the median $B M$ of triangle $A B C$ is equal to half the side $A C$, from which $\angle A B C=90^{\circ}$.

Let $B C=4 m$. Then $A C=6 m, A M=3 m, A B=\sqrt{36 m^{2}-16 m^{2}}=2 m \sqrt{5}$, and dividing $A M$ by $A B$, we get the answer.
3. In triangle $A B C$, point $M$ is the midpoint of $A C$, moreover, $B C=2 A C / 3$ and $\angle B M C=2 \angle A B M$. Find the ratio $A M / A B$.