Let $F$ be the point in which the parallel through $A$ to $BC$ intersects again the circumcircle of triangle $ABC$. We prove that the points $D$, $G$ and $F$ are collinear. $AFCB$ is a cyclic trapezoid, hence a cyclic one. If $T$ is the orthogonal projection of point $F$ onto $BC$, then $AFTD$ is a rectangle. It is easy to prove that triangles $ABD$ and $FCT$ are equal; it follows that $BD = CT$, i.e. $M$ is the midpoint of the line segment $DT$. It follows that $\frac{DM}{FA} = \frac{DM}{DT} = \frac{1}{2} = \frac{GM}{GA}$ and, since $\angle GMD = \angle GAF$, triangles $GMD$ and $GAF$ are similar. We deduce that $\angle DGM = \angle FGA$, i.e. points $D$, $G$, $F$ are collinear.

Then $\angle BED = \angle BEF = \angle BCF = \angle ABC = \angle ABD$, which shows that the line $AB$ is tangent to the circumcircle of triangle $BDE$.