Solution:

Let $F$ and $G$ be the points of intersection of line $BC$ with lines $AE$ and $AD$ respectively. The segment $CE$ is both bisector and altitude relative to side $AF$ of triangle $ACF$, hence it is also a median (and triangle $ACF$ is isosceles on base $AF$): we have $AE = EF$. Similarly, $BD$ is bisector and altitude (hence median) in triangle $ABG$, and $AD = DG$. Applying Thales' theorem in triangle $AFG$ gives the parallelism between lines $DE$ and $BC$.

Solution:

Let $I$ be the incenter of triangle $ABC$, let $BL$ and $CK$ be the bisectors from $B$ and from $C$, and let $\alpha, \beta, \gamma$ be the interior angles at $A, B, C$ respectively. Since angles $\widehat{AEI}$ and $\widehat{ADI}$ are right angles, the quadrilateral $AEID$ can be inscribed in a circle: we thus have $\widehat{IAE} = \widehat{IDE}$. On the other hand, $\widehat{AKC} = 180^{\circ} - \frac{\gamma}{2} - \alpha$, and (assuming that $E$ is interior to segment $IK$ in order to fix the configuration: the case in which this does not happen is analogous) $\widehat{KAE} = 90^{\circ} - \widehat{AKC} = \frac{\gamma}{2} + \alpha - 90^{\circ}$. The angles $\widehat{IAE}$ and $\widehat{IDE}$ are therefore equal to $\frac{\alpha}{2} - \widehat{KAE} = \frac{\alpha}{2} - \frac{\gamma}{2} - \alpha + 90^{\circ} = \frac{180^{\circ} - \gamma - \alpha}{2} = \frac{\beta}{2} = \widehat{LBC}$. The two alternate interior angles $\widehat{EDB}$ and $\widehat{DBC}$ formed by the transversal $DB$ with lines $ED$ and $BC$ are therefore congruent, and hence lines $ED$ and $BC$ are parallel.

Solution:

Let $D'$ be the projection of $D$ onto $BC$, and let $H$ be the foot of the altitude from $A$. Then
$$DD' = BD \sin \left(\frac{\beta}{2}\right) = AB \cos \left(\frac{\beta}{2}\right) \sin \left(\frac{\beta}{2}\right) = AB \sin \beta / 2 = AH / 2.$$
Similarly for $E$, hence $D$ and $E$ lie on the parallel to $BC$ located at distance $AH / 2$ on the side of $A$.