Let $G$ be the point such that $BCDG$ is a parallelogram and let $H$ be the midpoint of $AG$. Obviously $HEFD$ is also a parallelogram, and thus $DH = EF = l$. If $AD^2 + BC^2 = m^2$ is fixed, then from the Stewart theorem we have \[DH^2 =\frac{2DA^2 + 2DG^2 - AG^2}{4} = \frac{2m^2 - AG^2}{4},\] which is fixed. Thus $G$ and $H$ are fixed points, and from here the locus of $D$ is a circle with center $H$ and radius $l$. The locus of $B$ is the segment $(GI]$, where $I \in \Delta$ is a point in the positive direction such that $AI = a$. Finally, the locus of $C$ is a region of the plane consisting of a rectangle sandwiched between two semicircles of radius $l$ centered at points $H$ and $H'$, where $H'$ is a point such that $\stackrel{\longrightarrow}{IH'}=\stackrel{\longrightarrow}{GH}.$