Solution. Let the centers of $\Gamma_1$ and $\Gamma_2$ be $P$ and $Q$ respectively. Let the other intersection point of $\Gamma_2$ and $AB$ be $G$, the other intersection point of $\Gamma_1$ and $AC$ be $H$, the other tangent point of $B$ with respect to $\Gamma_2$ be $X_1$, and the tangent point of $C$ with respect to $\Gamma_1$ be $Y$, $Y_1$.
Lemma1. The three points $G, H, D$ are collinear and parallel to line $BC$.
Proof: We have $\angle AGD = \angle AFD = \angle AFC = \angle ABC$, so $GD \parallel BC$. Similarly $HD \parallel BC$, hence the three points $G, H, D$ are collinear.
Lemma2. $\triangle BXQ \sim \triangle CYP$
Proof: Since $\angle AQG = 2\angle ADG = 2\angle ADH = \angle APH$, $\triangle AQG$ is spirally similar to $\triangle APH$, so $\frac{AQ}{AP} = \frac{AG}{AH}$ and $\angle QAB = \angle QAG = \angle PAH = \angle PAC =$, and by Lemma1. $\frac{AG}{AH} = \frac{AB}{AC}$, we know $\triangle BAQ \sim \triangle CAP(\text{spirally similar})$, hence $\frac{BQ}{CP} = \frac{AQ}{AP} = \frac{XQ}{YP}$. Note that $\angle BXQ = \angle CYP = \frac{\pi}{2}$, so
$$\frac{BX}{CY} = \sqrt{\frac{BX^2}{CY^2}} = \sqrt{\frac{BQ^2 - QX^2}{CP^2 - PY^2}} = \sqrt{\frac{AQ^2}{AP^2}} = \frac{AQ}{AP},$$
and hence $\triangle BXQ \sim \triangle CYP(SSS)$.
---
## 2024-TWN — Page 85
Since $\triangle AQG$ is spirally similar to $\triangle APH$, $\triangle BXQ$ is spirally similar to $\triangle CYP$, or $\triangle BXQ$ is spirally similar to $\triangle CY_1P$. Without loss of generality, we have $\angle(BX, CY) = \angle(BQ, CP) = \angle(BA, CA)$, so the intersection point of $BX$ and $CY$ lies on $\Gamma$, namely $Z$.