Note that $2AM$ is actually $AB$. Moreover, since $AD=CD$ and $\angle ADB=\angle CND$, we have $$\triangle ABD\cong \triangle CND\iff AB=CN.\qquad (1)$$Supppse that $AB=CN$. We have $\triangle ABD\cong \triangle CND$ which implies that $BD=ND$. For convenience, denote $BD,DC$ by $x,y$, respectively. By Menelaus's Theorem, we have \begin{align*} \frac{BC}{DC}\cdot \frac{DN}{NA}\cdot\frac{AM}{MB}=1 &\implies\frac{x+y}{y}\cdot\frac{x}{y-x}=1 \\ &\implies x^2+2xy-y^2=0 \\ &\implies x=(\sqrt{2}-1)y \end{align*}Thus, $BC=x+y=\sqrt{2}y$. Also, $AC=\sqrt{2}\cdot DC=\sqrt{2}y$, as desired. Now, suppose that $CA=CB$. Since $CM$ bisects $AB$, we have $CM\perp AB$. Thus, $\angle BAD=\angle NCD$. Combining this with $AB=CD$ and $\angle ADB = \angle CND$, we have $\triangle ABD\cong \triangle CND$. Hence, from the result of $(1)$, we have $AB=CN$, as desired. Remark. This problem "only if" part is true only when $AC=BC$. $BC\neq BA$, otherwise in triangle $BAD$, there are $\angle ABD=\angle ADB=90^\circ$. While $AB=AC$ can happen, $CN=2AM$ is not necessary true.

SEE THE FIGURE:

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Notice if $CN=2AM=AB,$ then $\triangle ABD\cong\triangle CND$ by HL since $AD=CD.$ Then, $\angle MAN=\angle NCD$ so $\triangle AMN\sim\triangle CDN$ by AA. Hence, $\angle CMA=\angle CDN=90$ and $AC=BC.$ Conversely, suppose $AC=BC.$ Then, $\angle NMA=90$ so $\triangle AMN\sim\triangle CDN,$ which implies $\angle BAD=\angle NCD.$ Thus, $\triangle ABD\cong\triangle CND$ by ASA which implies $CN=AB.$ $\square$