jmc

We begin by drawing a diagram:

Since $D$ and $E$ are midpoints, $\overline{CD}$ and $\overline{BE}$ are medians. Let $F$ be the midpoint of $\overline{BC}$; we draw median $\overline{AF}$. The medians of a triangle are always concurrent (pass through the same point), so $\overline{AF}$ passes through $X$ as well.



The three medians cut triangle $ABC$ into six smaller triangles. These six smaller triangles all have the same area. (To see why, look at $\overline{BC}$ and notice that $△BXF$ and $△CXF$ have the same area since they share an altitude and have equal base lengths, and $△ABF$ and $△ACF$ have the same area for the same reason. Thus, $△ABX$ and $△ACX$ have the same area. We can repeat this argument with all three sizes of triangles built off the other two sides $\overline{AC}$ and $\overline{AB}$, to see that the six small triangles must all have the same area.)

Quadrilateral $AEXD$ is made up of two of these small triangles and triangle $BXC$ is made up of two of these small triangles as well. Hence they have the same area (and this will hold true no matter what type of triangle $△ABC$ is). Thus, the ratio of the area of quadrilateral $AEXD$ to the area of triangle $BXC$ is $1/1=\boxed{1}$.