Let $AM$ meet $\Gamma$ again at point $Y$, and let $XY$ meet $BC$ at point $D'$. By the same reasoning method, it suffices to prove that $D = D'$; we first prove the following lemma.

Lemma. Let the two diagonals of a cyclic quadrilateral $PQRS$ meet at point $T$, then
$$\frac{QT}{TS} = \frac{PQ \cdot QR}{PS \cdot SR}.$$

Proof of Lemma. Denote the (signed) area of triangle $W_1W_2W_3$ by $[W_1W_2W_3]$, then
$$\frac{QT}{TS} = \frac{[PQR]}{[PSR]} = \frac{\frac{1}{2}PQ \cdot QR \sin \angle PQR}{\frac{1}{2}PS \cdot SR \sin \angle PSR} = \frac{PQ \cdot QR}{PS \cdot SR}$$
(since $\angle PQR$ and $\angle PSR$ are supplementary). This proves the lemma.

Applying the above lemma to quadrilaterals $ABYC$ and $XBYC$ respectively, we know
$$1 = \frac{BM}{MC} = \frac{AB \cdot BY}{AC \cdot CY} \quad \text{and} \quad \frac{BD'}{D'C} = \frac{XB \cdot BY}{XC \cdot CY}.$$
Combining these two equations, we get
$$\frac{BD'}{D'C} = \frac{XB \cdot BY}{XC \cdot CY} = \frac{XB \cdot AC}{XC \cdot AB}. \quad (1)$$

Hence triangle $XBF$ is similar to $XCE$, and we obtain
$$\frac{XB}{XC} = \frac{BF}{CE}. \qquad (2)$$

Since $\angle FIB = \angle AIB - 90^\circ = \frac{1}{2} \angle ACB = \angle ICB$ and $\angle FBI = \angle IBC$,
we get that triangle $FBI$ is similar to $IBC$. Similarly, triangle $EIC$ is also similar to $IBC$. Hence
$$\frac{FB}{IB} = \frac{BI}{BC} \quad \text{and} \quad \frac{EC}{IC} = \frac{IC}{BC}. \qquad (3)$$

Now draw a line parallel to $BC$ and tangent to the incircle, and let it meet sides $AB$ and $AC$ at points $B_1, C_1$ respectively. Also let the incircle be tangent to sides $AB, AC$ at points $B_2, C_2$ respectively. By the homothety, line $B_1I$ is parallel to the external angle bisector of $\angle ABC$, so $\angle B_1IB = 90^\circ$. Also $\angle BB_2I = 90^\circ$, from which we know $BB_2 \cdot BB_1 = BI^2$. Similarly we know $CC_2 \cdot CC_1 = CI^2$. Therefore
$$\frac{BI^2}{CI^2} = \frac{BB_2 \cdot BB_1}{CC_2 \cdot CC_1} = \frac{BB_1}{CC_1} \cdot \frac{BD}{CD} = \frac{AB}{AC} \cdot \frac{BD}{CD}. \quad (4)$$

Combining Eq. (1), Eq. (2), Eq. (3) and Eq. (4), we get
$$\frac{BD'}{CD'} = \frac{XB}{XC} \cdot \frac{AC}{AB} = \frac{BF}{CE} \cdot \frac{AC}{AB} = \frac{BI^2}{CI^2} \cdot \frac{AC}{AB} = \frac{BD}{CD},$$

hence $D = D'$. This completes the proof.