Solution. Since $\angle ADF$ is the angle between $\Gamma$ and $\odot(ADE)$, consider the circle $\odot(ADE)$ and its second intersection $A'$ with $\odot(ABC)$. Also let $DA'$ intersects $BX$ at $X'$. Denote $\angle ADF$ by $\theta$. Now we first show the following two lemmas.

Lemma 1. $(A, A'; B, C) = \tan^2(\frac{1}{2}\theta)$ or $\cot^2(\frac{1}{2}\theta)$.

Proof. Note that $\odot(ABC)$ is orthogonal to both $\Gamma$ and $\odot(ADE)$ as $OA^2 = OC^2 = OD \cdot OE$. Therefore, if we take the inversion at $D$ and denote the image of a point $P$ by $P_*$, then $A_*A_*'$, $B_*C_*$ are diameters of a circle that intersect with angle $\theta$. Thus it is clear that
$$(A, A'; B, C) = (A_*, A_*'; B_*, C_*) = \tan^2(\frac{1}{2}\theta) \text{ or } \cot^2(\frac{1}{2}\theta).$$

Lemma 2. $DX \perp DX'$.

Proof. Take the inversion at $C$ with radius $CB = CD$. This inversion sends $\odot(ABC)$ to a line passing through $B$ and perpendicular to $CO$, and thus it is $BXX'$. This shows that the inversion sends $A$ to $X$ and $A'$ to $X'$. Moreover, the inversions sends $\Gamma$ to $BD$, and so $E$ is sent to the reflection $D'$ of $D$ with respect to $B$ as $(E, D; B, C)$ is harmonic. Since $XX' \perp DD'$, we have that $XX'$ is the perpendicular bisector of $DD'$. Moreover, since $A, A', D, E$ are concyclic, we have that $X, X', D, D'$ are also concyclic. As a consequence, $DX \perp DX'$.

To show the original statement, note that by perspectivity through $C$, we have
$$\tan^2(\frac{1}{2}\theta) \text{ or } \cot^2(\frac{1}{2}\theta) = (A, A'; B, C) = \frac{XB}{BX'} = \cot^2 \angle BDX$$
where the first equality follows from the first lemma and the last equality follows from the second lemma. Thus either $\theta/2 = BDX$ or $\theta/2 = 90^\circ - \angle BDX = \angle BXD$, as desired.