13. (16 points) Let the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ pass through the point $(0, \sqrt{3})$, with an eccentricity of $\frac{1}{2}$. The line $l$ passes through the right focus $F$ of the ellipse $C$, intersecting the ellipse $C$ at points $A$ and $B$. The projections of $A$, $F$, and $B$ on the line $x=4$ are $D$, $K$, and $E$ respectively.
(1) Find the equation of the ellipse $C$.
(2) Connect $A E$ and $B D$. When the inclination angle of the line $l$ changes, do the lines $A E$ and $B D$ intersect at a fixed point? If so, find the coordinates of the fixed point and provide a proof; otherwise, explain the reason.
(1) According to the ellipse passing through the point $(0, \sqrt{3})$, we know $b=\sqrt{3}$.
Then, from $e=\frac{c}{a}=\frac{\sqrt{a^{2}-b^{2}}}{a}=\frac{1}{2} \Rightarrow a=2$.
Therefore, the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(2) When the slope of the line $l$ does not exist, the line $l \perp$ $x$-axis, then the quadrilateral $A B E D$ is a rectangle.

By symmetry, we know that the line $A E$ intersects with $B D$ at the midpoint $N\left(\frac{5}{2}, 0\right)$ of $F K$.

Below is the proof: When the slope of the line $l$ exists, the line $A E$ intersects with $B D$ at the fixed point $N\left(\frac{5}{2}, 0\right)$.
As shown in Figure 2.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), D\left(4, y_{1}\right)$,
$$
E\left(4, y_{2}\right), l_{A B}: y=k(x-1) \text {. }
$$

By combining the equation of the ellipse $C$, and eliminating $y$, we get the equation about $x$
$$
\begin{array}{l}
\left(3+4 k^{2}\right) x^{2}-8 k^{2} x+4 k^{2}-12=0 \\
\Rightarrow x_{1}+x_{2}=\frac{8 k^{2}}{3+4 k^{2}}, x_{1} x_{2}=\frac{4 k^{2}-12}{3+4 k^{2}} .
\end{array}
$$

Notice that, $l_{A E}: y-y_{2}=\frac{y_{2}-y_{1}}{4-x_{1}}(x-4)$.
When $x=\frac{5}{2}$,
$$
y=k\left(x_{2}-1\right)-\frac{k\left(x_{2}-x_{1}\right)}{4-x_{1}} \cdot \frac{3}{2}
$$
$$
\begin{array}{l}
=\frac{5 k\left(x_{1}+x_{2}\right)-2 k x_{1} x_{2}-8 k}{2\left(4-x_{1}\right)} \\
=\frac{-8 k\left(3+4 k^{2}\right)-2 k\left(4 k^{2}-12\right)+5 k \cdot 8 k^{2}}{2\left(4-x_{1}\right)\left(3+4 k^{2}\right)} \\
=0,
\end{array}
$$

That is, the point $N\left(\frac{5}{2}, 0\right)$ lies on the line $A E$.
Similarly, the point $N\left(\frac{5}{2}, 0\right)$ lies on the line $B D$.
Thus, when the inclination angle of the line $l$ changes, the line $A E$ intersects with $B D$ at the fixed point $N\left(\frac{5}{2}, 0\right)$.