As the $A$-bisector is also the altitude, we have $BD \perp AD$. Let $I$ be the incenter of $ABC$ and $E' \neq B$ the intersection of $BI$ and the circumcircle of $ABD$ (Fig. 2). As the angles $\angle ABE' = \angle DBE'$ correspond to the arcs $E'A$ and $E'D$ of this circle, we have $E'A = E'D$. Thus $E'$ lies on the perpendicular bisector of $AD$, meaning $E' = E$. Hence $ABDE$ is a cyclic quadrilateral (Fig. 3).
It remains to verify that $\angle EDI + \angle CID = 90^\circ$, which would yield $CI \perp DE$, as desired. To show this, let $\angle ACB = \angle ABC = \gamma$. Then
$$\angle EDI = \angle EDA = \angle EBA = \frac{\gamma}{2},$$
whereas
$$\angle CID = 180^\circ - \angle CDI - \angle DCI = 180^\circ - 90^\circ - \frac{\gamma}{2} = 90^\circ - \frac{\gamma}{2}.$$
Thus $\angle EDI + \angle CID = 90^\circ$, finishing the proof.

As the $A$-bisector is also the altitude, we have $BD \perp AD$. Thus the perpendicular bisector of $AD$ is parallel to $BC$, therefore containing the midline of $ABC$ and bisecting $AB$ and $AC$. Let $L$ be the midpoint of $AB$ and $I$ the incenter of $ABC$ (Fig. 4). Then $\angle LEB = \angle CBE = \angle LBE$, meaning that $LBE$ is isosceles and $LE = LB = LA$. Thus $L$ is the circumcenter of $ABE$ and $AB$ is a diameter of the circle. Now as $ADB = 90^{\circ}$, the point $D$ also lies on this circle, meaning $ABDE$ is cyclic. We proceed like in Solution 1.

As the $A$-bisector is also the altitude, we have $BD \perp AD$. Thus the perpendicular bisector of $AD$ is parallel to $BC$, therefore containing the midline of $ABC$ and bisecting $AB$ and $AC$. Let $L$ and $M$ be the midpoints of $AB$ and $AC$, respectively. Then $\angle LEB = \angle CBE = \angle LBE$, meaning that $LBE$ is isosceles and $LE = LB = LA$. Also $MC = MA = LB = LE$, and as $LM$ is the midline of $ABC$, we have $LM = \frac{1}{2}BC = CD$.
Let $N$ be the intersection of $DE$ and $AC$; depending on the orientation of points (Figures 5 and 6) we have
$$CN = CM \pm MN, \\ CD = LM = LE \pm EM = CM \pm ME.$$
Triangles $NCD$ and $NME$ are similar by parallel sides, giving $\frac{CD}{ME} = \frac{CN}{MN}$. Substituting in the previous equation, we get $\frac{CM \pm ME}{ME} = \frac{CM \pm MN}{MN}$, from which $\frac{CM}{ME} \pm 1 = \frac{CM}{MN} \pm 1$, which in turn implies $\frac{CM}{ME} = \frac{CM}{MN}$. Therefore $ME = MN$, from which $CD = CN$. The claim of the problem follows, as the $C$-bisector in the isosceles triangle $CDN$ and the side $DN$ are perpendicular.

Fig. 5

Fig. 6

Let $I$ be the incenter of $ABC$. Clearly $D$ is the point of tangency of the incircle and side $BC$. Let $N$ be the tangency point with side $AC$ (Fig. 7), then $CD = CN$. In the isosceles triangle $CDN$, the line $CI$ is both the angle bisector and altitude, so $DN$ and $CI$ are perpendicular. It remains to show that $E$ lies on $DN$.
By the Iran lemma, the angle bisector $BI$ and the extensions of the incircle chord $DN$ and the midline corresponding to $BC$ intersect at one point. Of those, the first and the third intersect at $E$, so $E$ must lie on $DN$, as desired.

Fig. 7