Three. (20 points) Let $\Gamma$ be the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0), A(x, y)$ be a point on $\Gamma$, and $B, C, D$ be the points symmetric to $A$ with respect to the $y$-axis, the origin, and the $x$-axis, respectively. Let $E$ be a point on $\Gamma$ such that $A E \perp A C$, and let $F$ be the intersection of $C E$ and $B D$.
(1) Find the coordinates of point $F$ (expressed in terms of the coordinates of $A$);
(2) As $A$ moves along $\Gamma$, what is the trajectory of $F$? What is its relationship with $\Gamma$?
Three, let $E$ be $\left(x_{1}, y_{1}\right)$, from $k_{A C} \times k_{A E}=-1$, we have
$$
\frac{y}{x} \cdot \frac{y_{1}-y}{x_{1}-x}=-1 \text {. }
$$

Rearranging gives $y_{1}=\frac{x^{2}+y^{2}}{y}-\frac{x}{y} \cdot x_{1}$.
Also, $\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=1$, substituting $y_{1}$ into equation (1) yields
$$
\left(x-x_{1}\right)\left[a^{2} x^{3}+2 a^{2} y^{2} x-b^{2} y^{2} x-\left(a^{2} x^{2}+b^{2} y^{2}\right) x_{1}\right]=0 \text {. }
$$

Thus, $x_{1}=\frac{a^{2} x^{3}+2 a^{2} y^{2} x-b^{2} y^{2} x}{a^{2} x^{2}+b^{2} y^{2}}$
$$
=\left(1+2 y^{2} \cdot \frac{a^{2}-b^{2}}{a^{2} x^{2}+b^{2} y^{2}}\right) x \text {. }
$$

Similarly, $y_{1}=\left(1-2 x^{2} \cdot \frac{a^{2}-b^{2}}{a^{2} x^{2}+b^{2} y^{2}}\right) y$.
Let $F(t x,-t y)$, then $\frac{t x+x}{-t y+y}=\frac{x_{1}+x}{y_{1}+y}$. Therefore,
$$
t=\frac{x_{1} y-x y_{1}}{x_{1} y+x y_{1}+2 x y}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}} \text {. }
$$

Thus, $F\left(\frac{a^{2}-b^{2}}{a^{2}+b^{2}} \cdot x, \frac{a^{2}-b^{2}}{a^{2}+b^{2}} \cdot(-y)\right)$.
When $A(x, y)$ moves along $\Gamma$, then $F\left(x^{\prime}, y^{\prime}\right)$ satisfies
$$
\frac{x^{\prime 2}}{\left(\frac{a^{2}-b^{2}}{a^{2}+b^{2}} \cdot a\right)^{2}}+\frac{y^{\prime 2}}{\left(\frac{a^{2}-b^{2}}{a^{2}+b^{2}} \cdot b\right)^{2}}=1 \text {. }
$$

Therefore, the trajectory of $F$ is an ellipse, and it is similar to $\Gamma$.