6. As shown in Figure 3, circles $\odot B$ and $\odot C$ with radii 3 and 5, respectively, are externally tangent at point $A$. One of their external common tangents touches $\odot B$ at point $D$ and $\odot C$ at point $E$. A perpendicular line to $DE$ through point $A$ intersects the perpendicular bisector of $BC$ at point $F, H$ is the midpoint of $BC$. Then the area of $\triangle AHF$ is ( ).
(A) $\frac{\sqrt{11}}{2}$
(B) $\frac{\sqrt{13}}{2}$
(C) $\frac{\sqrt{15}}{2}$
(D) $\frac{\sqrt{17}}{2}$
6. C.

As shown in Figure 7, connect $BD$, $CE$, and draw a perpendicular from point $B$ to $CE$, with the foot of the perpendicular being $M$.
Since $DE$ is the common tangent of the two circles, and $D$, $E$ are the points of tangency, we know $BD \perp DE$, $CE \perp DE$
$\Rightarrow$ Quadrilateral $BDEM$ is a rectangle
$$
\begin{array}{l}
\Rightarrow BM // DE, BD = EM \\
\Rightarrow CM = CE - BD = 2 .
\end{array}
$$

Then $\triangle F H A \sim \triangle B M C$
$$
\Rightarrow \frac{S_{\triangle A H F}}{S_{\triangle C M B}}=\left(\frac{A H}{C M}\right)^{2} \text {. }
$$

Also, $A H = B H - B A = \frac{B A + A C}{2} - B A = 1$, then
In the right triangle $\triangle B C M$, according to the Pythagorean theorem, we have
$$
\begin{array}{l}
B M = \sqrt{8^{2} - 2^{2}} = 2 \sqrt{15} . \\
\text { Therefore, } S_{\triangle C M B} = \frac{1}{2} \times 2 \sqrt{15} \times 2 = 2 \sqrt{15} \\
\Rightarrow S_{\triangle A H F} = S_{\triangle C M B} \left(\frac{1}{2}\right)^{2} = \frac{\sqrt{15}}{2} .
\end{array}
$$