Three, Answer the following questions (15 points each, total 30 points, detailed process required)
13. (15 points) In the figure, the two legs $A B$ and $B C$ of the right triangle $A B C$ have lengths of 3 and 4, respectively. The triangle $A B C$ is rotated clockwise around point $C$ to $A_{1} B_{1} C$, such that $A_{1} C$ and $B_{1} C$ lie on line $l$. $A_{1} A$ intersects $B_{1} C$ at $D$. Find $\frac{\mathrm{AD}}{\mathrm{A}_{1} \mathrm{D}}$.
【Analysis】As shown in the figure, construct the triangle $A_2BC$ symmetrical to triangle $ABC$ about $A_1B$, and connect $AB_1$, $A_2B$, and $A_2C$. Based on the relationships between the angles, deduce that points $A_2$, $C$, and $B_1$ are collinear. Then, use the relationship that the area of a triangle is proportional to its base to solve the problem.

【Solution】Solution: As shown in the figure, construct the triangle $A_2BC$ symmetrical to triangle $ABC$ about $A_1B$, and connect $AB_1$, $A_2B$, and $A_2C$. Since $AB=3$, $BC=4$, it follows that $A_2C=5$, $B_1C=4$, and $S_{\triangle AA_2C}=2S_{\triangle ABC}=12$.

Since $\angle ACB + \angle ACB_1 + \angle A_1CB_1 = \angle ACB + \angle ACB_1 + \angle A_2CB = 180^\circ$,
points $A_2$, $C$, and $B_1$ are collinear, and
$\frac{S_{\triangle AA_2C}}{S_{\triangle B_1CC}} = \frac{A_2C}{B_1C} = \frac{5}{4}$, which means $S_{\triangle AB_1C} = \frac{48}{5}$.
Since $\frac{S_{\triangle ACD}}{S_{\triangle CDA_1}} = \frac{AD}{A_1D}$, and $\frac{S_{\triangle A_1B_1}}{S_{\triangle B_1A_1D}} = \frac{AD}{A_1D}$,
it follows that $\frac{S_{\triangle ACD} + S_{\triangle ADB_1}}{S_{\triangle CDA_1} + S_{\triangle B_1A_1D}} = \frac{AD}{A_1D}$.
Therefore, $\frac{AD}{A_1D} = \frac{48}{5} \times \frac{1}{6} = \frac{8}{5}$.