Note that $\angle NEA=\angle BAD=\angle DCB$ and $\angle ANE=180^\circ-\angle BAC=\angle BDC$, so $\triangle NEA \sim \triangle DCB$. Thus, $\triangle BDM\sim \triangle ADE$, so $D$ spirals $BA$ to $ME$. This means that $\angle DEM=\angle DAB=\angle DCB$, as desired.

Complex bash with $\triangle ABC$ inscribed in the unit circle, so that $$|a|=|b|=|c|=|d|=1$$$$m=\frac{b+c}2$$$$n=\frac{a+d}2$$Now we find the coordinate of $E$. Since $E$ lies on line $AC$, we have $$\overline{e} = \frac{a+c-e}{ac}$$Since the circumcircle of $\triangle AEN$ is tangent to $AB$ at $A$, we have $$\frac{(a-e)(a-n)}{(a-b)(e-n)} \in \mathbb{R} \implies \frac{(a-c)(a-d)}{(a-b)(a+d-2e)} \in \mathbb{R}$$$$\frac{(a-c)(a-d)}{(a-b)(a+d-2e)} = -\frac{b(a-c)(a-d)}{c(a-b)(a+d-2ad\overline{e})} = -\frac{b(a-c)(a-d)}{(a-b)(ac-cd-2ad+2de)}$$$$ac-cd-2ad+2de = -b(a+d-2e)$$$$ac-cd-2ad+2de+ab+bd-2be = 0$$$$e = \frac{ab+ac-2ad+bd-cd}{2(b-d)}$$Then we have the vectors \begin{align*} m-c &= \frac{b-c}2 \\ e-d &= \frac{(a-d)(b+c-2d)}{2(b-d)} \\ e-m &= \frac{(a-b)(b+c-2d)}{2(b-d)} \end{align*}Thus $$\frac{(m-c)(e-d)}{(c-d)(e-m)} = \frac{(a-d)(b-c)}{2(a-b)(c-d)} \in \mathbb{R}$$as desired. $\blacksquare$