The lines $BC$ and $DE$ are parallel, so the angles $BED$ and $DAE$ are equal. Then so are the angles $AED$ and $DBE$. Let $I$ and $J$ be the incentres of the triangles $ADE$ and $BDE$, respectively. It follows that the triangles $DIE$ and $DJB$ are similar, so $DI/DE = DJ/DB$. Since the angles $IDJ$ and $EDB$ are equal, the triangles $DIJ$ and $DEB$ are similar, so the angles $DIJ$ and $DEB$ are equal. Let the line $BE$ meet the line $IJ$ at the point $F$. Notice that the quadrangle $DIEF$ is cyclic to deduce that the angles $EFI$ and $EDI$ are both equal to one half of the angle $ACB$. Consequently, the line $IJ$ is parallel to the exterior bisectrix of the angle $AEB$. The conclusion follows.