IRL ABooklet

Extend $AD$ and $AE$ to meet the circumcircle of $\triangle ABC$ at $F$ and $G$, respectively. Let $O$ be the intersection point of $BG$ and $CF$.

Since $\angle BAD = \angle CAE$, we have
$$\angle BCO = \angle OGF = \angle BAD = \angle CAE = \angle CBO = \angle OFG,$$
subtending equal arcs. Hence $OB$ and $OC$ are tangents to the circumcircles of $\triangle BAD$ and $\triangle CAE$ by the Alternate Segment Theorem, and $\triangle OBC$ is isosceles with $|OB| = |OC|$. Then $O$ must be on the common tangent to the circumcircles of $\triangle BAD$ and $\triangle CAE$, since it has the same power with respect to them. Since $OA$ is tangent to the two circles, $|OA| = |OB| = |OC|$ hence $O$ is the circumcentre of $\triangle ABC$, hence $BG$ and $CF$ are diameters, thus $\angle BAE = \angle CAD = 90^\circ$.

Let $M, N$ be the centres of the circumcircles of triangles $BAD$ and $CAE$, respectively. These two circles are tangent at $A$ iff $M, A, N$ are collinear. We define $\alpha = \angle BAD = \angle CAE$, $\beta = \angle MAB = \angle MBA$, and $\gamma = \angle NAC = \angle NCA$.

We then have $\angle BMA = 180^\circ - 2\beta$ and $\angle ANC = 180^\circ - 2\gamma$. If $M$ and $D$ are on opposite sides of $AB$, we have $\angle BDA = 180^\circ - \frac{1}{2}\angle BMA = 90^\circ + \beta$ and $\angle ADE = 180^\circ - \angle BDA = 90^\circ - \beta$. Similarly, $\angle AED = 90^\circ - \gamma$. We now obtain
$$\angle DAE = 180^\circ - \angle ADE - \angle AED = \beta + \gamma.$$
Because $M, A, N$ are collinear, this implies $2\alpha + 2\beta + 2\gamma = 180^\circ$, hence $\alpha + \beta + \gamma = 90^\circ$. Finally, $\angle BAE = \angle BAD + \angle DAE = \alpha + \beta + \gamma = 90^\circ$ and $\angle DAC = \angle DAE + \angle EAC = \beta + \gamma + \alpha = 90^\circ$.

If $M$ and $D$ were on the same side of $AB$, the above argument, with $\beta$ replaced by $-\beta$ in the formulas for $\angle BDA$, $\angle ADE$ etc., gives the same result. We then see that $\angle BDA = 90^\circ + \alpha > 90^\circ$, which implies that $M$ and $D$ must actually be on opposite sides of $AB$.