6. There is a triangular activity site, with side $A B=20, \angle B=60^{\circ}, \angle C=75^{\circ}, A D$ is a lime line, and $A D \perp B C$ at point $D$. Xiao Ming starts from point $D$, passes through a point on side $A B$ and a point on side $A C$, and finally returns to $D$. Then the shortest distance Xiao Ming walks is ( ).
(A) $5 \sqrt{6}$
(B) $10 \sqrt{3}$
(C) $10 \sqrt{6}$
(D) $20 \sqrt{6}$
6. C.

As shown in Figure 7, construct the symmetric point $D_{1}$ of point $D$ with respect to $AB$, and the symmetric point $D_{2}$ of point $D$ with respect to $AC$. Connect $D_{1}D_{2}$, which intersects $AB$ and $AC$ at points $E$ and $F$, respectively. Then $DE + EF + FD$ is the shortest path that Xiao Ming can take.

The reason is as follows: Take any point $E'$ on $AB$ and any point $F'$ on $AC$, and connect $DE'$, $E'F'$, $F'D$, $E'D_{1}$, $F'D_{2}$, $AD_{1}$, and $AD_{2}$. It is easy to see that
$$
\begin{array}{l}
E'D = E'D_{1}, ED = ED_{1}, \\
F'D = F'D_{2}, FD = FD_{2}.
\end{array}
$$

Therefore, $DE + EF + FD = D_{1}E + EF + FD_{2}$
$$
\begin{array}{l}
= D_{1}D_{2}, \\
DE' + E'F' + F'D = D_{1}E' + E'F' + F'D_{2}.
\end{array}
$$

According to the principle that "the shortest distance between two points is a straight line," we have
$$
D_{1}D_{2} \leq D_{1}E' + E'F' + F'D_{2}
$$
(when $E'$ coincides with $E$ and $F'$ coincides with $F$, equality holds), i.e.,
$$
DE + EF + FD \leq DE' + E'F' + F'D.
$$

Therefore, $DE + EF + FD$ is the shortest path that Xiao Ming can take.

Since $AD = AB \sin B = 20 \sin 60^{\circ} = 10 \sqrt{3}$, we have $AD_{1} = AD_{2} = AD = 10 \sqrt{3}$.
$$
\text{Also, } \angle D_{1}AB = \angle DAB, \angle D_{2}AC = \angle DAC,
$$
then
$$
\begin{array}{l}
\angle D_{1}AD_{2} = 2 \angle BAC \\
= 2(180^{\circ} - \angle B - \angle C) = 90^{\circ}.
\end{array}
$$

By the Pythagorean theorem, we get
$$
D_{1}D_{2} = \sqrt{AD_{1}^2 + AD_{2}^2} = \sqrt{(10 \sqrt{3})^2 \times 2} = 10 \sqrt{6}.
$$

Therefore, $DE + EF + FD = 10 \sqrt{6}$.