Um this is doable paperless. Note that $D$ is midpoint of $AB$, and $DM = DN$, which implies $AN \parallel BM$. So, \[ \measuredangle BAC = \measuredangle BMC = \measuredangle (MB, MC) = \measuredangle (AN,MC) = \measuredangle AND = \measuredangle AED = \measuredangle ACB \]

My solution during contest: [asy][asy] usepackage("tikz"); label("\begin{tikzpicture} \coordinate[label=above right:$A$] (A) at (1.02,3.87); \coordinate[label=right:$C$] (C) at (3.88,-.96); \coordinate[label=left:$B$] (B) at (-3.44,-2.04); \coordinate[label=below:$N$] (N) at (1.16,.04); \coordinate[label=above:$D$] (D) at (-1.21,.91); \coordinate[label=left:$M$] (M) at (-3.58,1.78); \draw[thick] (A)--(B)--(C)--cycle; \draw[thick] (0,0) circle (4); \draw[thick,red] (C)--(M); \coordinate[label=right:$E$] (E) at (2.45,1.46); \draw[thick,blue] (.51,1.93) circle (2); \foreach \s in {A,B,C,D,E,N,M}\filldraw (\s) circle (1.5pt); \end{tikzpicture}"); [/asy][/asy] Let $AD=DB=x,AE=EC=y,DN=DM=z,$ and $CN=m$. By PoP, we have \begin{align*} CD \cdot DM= BD \cdot DA &\Longrightarrow z^2+mz &= x^2\\ CN\cdot CD = CE \cdot CA &\Longrightarrow m^2+mz &= 2y^2. \end{align*}We have $(m+z)^2=x^2+2y^2$. By LoC of $\triangle ACD$: \[\cos A = \frac{x^2 + 4y^2 - (m+z)^2}{4xy}=\frac{2y^2}{4xy}=\frac{y}{2x}.\]By LoC of $\triangle ABC$: \[BC^2 = 4x^2+4y^2- 8xy\cos A = 4x^2 + 4y^2-4y^2=4x^2 = AB^2,\]so we have $BC=AB$. $\blacksquare$

Let me complete the statement GorgonMathDota wrote: Um this is doable paperless and within 15 seconds Notice that $MBNA$ is a parallelogram so $$\measuredangle BCA= \measuredangle DEA = \measuredangle DNA = \measuredangle NMB = \measuredangle CMB = \measuredangle CAB \ \blacksquare$$

Note that $MANB$ is a parallelogram since its diagonals bisect each other. $$\angle{ACB}=\angle{AED}=\angle{AND}=\angle{NMB}=\angle{CAB}$$and we are done.

Clearly, $BNAM$ is a parallelogram.$$\angle BAC=\angle BMC=\angle AND=\angle DEA=\angle BCA \implies QED$$