jmc

Since $G$ and $H$ are midpoints, we know that $DG=GC$ and $EH=HF.$ From vertical angles, we can see that $\angle DHE\equiv \angle FHJ.$ Finally, from parallel lines, it is clear that $\angle DEH\equiv \angle HFJ.$ We have now found two angles and a side equal in triangles $DEH$ and $JFH,$ so therefore, $△DEH\equiv △JFH.$ Looking at areas, we have: $$\begin{array}{rl}[CDEF]& =[CDHF]+[DEH]\\ [CDJ]& =[CDHF]+[HFJ]\end{array}$$ However, we just proved that $△DEH\equiv △JFH,$ and so $[HFJ]=[DEH].$ Therefore,$$[CDEF]=[CDJ]=\boxed{36}.$$