13. Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ passes through the point $(0, \sqrt{3})$, and the eccentricity is $\frac{1}{2}$. The line $l$ passes through the right focus $F$ of the ellipse $C$ and intersects the ellipse at points $A, B$. The projections of points $A, F, B$ on the line $x=4$ are $D, K, E$ respectively.
(1) Find the equation of the ellipse $C$;
(2) Connect $A E$ and $B D$. Explore whether the lines $A E$ and $B D$ intersect at a fixed point when the inclination angle of line $l$ changes. If so, find the coordinates of the fixed point and provide a proof; if not, explain the reason.
(1) $b=\sqrt{3}, e=\frac{c}{a}=\frac{1}{2} \Rightarrow a=2, c=1$, so the equation of the ellipse $C$ is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(2) As shown in the figure, let $A B: x=n y+1, A\left(x_{1}, y_{1}\right)$,
$B\left(x_{2}, y_{2}\right), E\left(4, y_{2}\right)$, combining $\left\{\begin{array}{l}x=n y+1, \\ 3 x^{2}+4 y^{2}=12\end{array}\right.$
$$
\Rightarrow\left(3 n^{2}+4\right) y^{2}+6 n y-9=0 \text {. }
$$

Then $A E: y-y_{2}=\frac{y_{2}-y_{1}}{4-x_{1}}(x-4)$,
Let $y=0 \Rightarrow x_{P}=4-\frac{y_{2}\left(4-x_{1}\right)}{y_{2}-y_{1}}=\frac{x_{1} y_{2}-4 y_{1}}{y_{2}-y_{1}}=\frac{n y_{1} y_{2}+y_{2}-4 y_{1}}{y_{2}-y_{1}}$,
By Vieta's formulas, $y_{1}+y_{2}=-\frac{6 n}{3 n^{2}+4}, y_{1} y_{2}=-\frac{9}{3 n^{2}+4} \Rightarrow n y_{1} y_{2}=\frac{3}{2}\left(y_{1}+y_{2}\right)$,
Substituting into the above equation, we get $x_{P}=\frac{5}{2}$, similarly, we can get that $B D$ intersects the $x$-axis at point $P$.
Therefore, the lines $A E$ and $B D$ intersect at the fixed point $P\left(\frac{5}{2}, 0\right)$.