Suppose $ABCD$ is a parallelogram and $E$ is a point between $B$ and $C$ on the line $BC$.  If the triangles $DEC$, $BED$ and $BAD$ are isosceles what are the possible values for the angle $DAB$?
1. **Identify the given information and draw the parallelogram:**
   - $ABCD$ is a parallelogram.
   - $E$ is a point on $BC$ such that $B$ is between $B$ and $C$.
   - Triangles $DEC$, $BED$, and $BAD$ are isosceles.

2. **Analyze the isosceles triangles:**
   - Since $\triangle DEC$ is isosceles, $DE = EC$ or $DE = DC$.
   - Since $\triangle BED$ is isosceles, $BE = ED$ or $BE = BD$.
   - Since $\triangle BAD$ is isosceles, $BA = AD$.

3. **Use the properties of the parallelogram:**
   - Opposite sides of a parallelogram are equal, so $AB = CD$ and $AD = BC$.
   - Opposite angles of a parallelogram are equal, so $\angle DAB = \angle BCD$ and $\angle ABC = \angle CDA$.

4. **Consider the isosceles triangles and their implications:**
   - For $\triangle DEC$ to be isosceles, $DE = EC$ implies $E$ is the midpoint of $DC$.
   - For $\triangle BED$ to be isosceles, $BE = ED$ implies $E$ is the midpoint of $BC$.
   - For $\triangle BAD$ to be isosceles, $BA = AD$ implies $\angle DAB = \angle BDA$.

5. **Determine the possible values for $\angle DAB$:**
   - Since $E$ is the midpoint of $BC$, $E$ divides $BC$ into two equal segments.
   - Given that $ABCD$ is a parallelogram, $\angle ABC = \angle CDA$.
   - If $m(\angle B) = 135^\circ$, then $\angle DAB = 45^\circ$ because $\angle DAB + \angle ABC = 180^\circ$ in a parallelogram.

6. **Verify the hypothesis:**
   - If $AB \perp AC$, then $\angle BAC = 90^\circ$.
   - Since $X$ is the midpoint of $BC$, $E$ is also the midpoint of $BC$.
   - The hypothesis is accomplished if $m(\angle B) = 135^\circ$.

The final answer is $\boxed{45^\circ}$