Let $E$ be reflection of $A$ in perpendicular bisector of $BC$; easily $ABCE$ is an isosceles trapezoid with $CE=AE=AB$; if $F$ is projection of $E$ onto $BC$, $ADFE$ is a rectangle, hence $AB=AE=DF=2DM$, done. Best regards, sunken rock

Construct angle bisector $BN$ such that $\angle NBC=\angle NCB=\angle B$. $NBC$ is a isosceles triangle so $NM$ is perpendicular bisector of $\angle BNC$ and $NM \parallel AD $. By thales theorem $AN/NC=DM/MC$ And by angle bisector theorem $AN/NC=AB/BC$ So ultimately $AB/BC=DM/MC$ or $AB=2DM$. so done

Let $AB=x \text{ and } \angle B = 2y= 2\angle C$; Hence,$$BD=x\cos 2y \text{ and } DC=2x\cos ^2y \implies BM= \frac{x}{2}[\cos 2y +2\cos ^2y] \implies DM=\frac{x}{2}[\cos 2y +2\cos ^2y] - x\cos 2y = \frac{x(2\cos ^2y -2\cos ^2y +1)}{2}=\frac{x}{2}=\frac{1}{2}AB \implies \boxed{AB=2DM}$$

Let $N$ bet middle of $AB$. Let $\angle NBD=2x, \angle BCD=x$. Then $\angle BDN=2x$,this why $\angle NDA=90-2x$.We must prove that $ND=DM$, or $\angle DMN=x$. We know that $NM||AC$, this why $,\angle DMN=\angle DNM =x$, and $2DM=AB$

Midline !!