8. (10 points) Five points $A$, $B$, $C$, $D$, $E$ on a plane satisfy: $A B=8$ cm, $B C=4$ cm, $A D=5$ cm, $D E=1$ cm, $A C=12$ cm, $A E=6$ cm. If the area of triangle $E A B$ is 24 square cm. Then the distance from point $A$ to $C D$ is . $\qquad$
8. (10 points) Five points $A, B, C, D, E$ on a plane satisfy: $A B=8$ cm, $B C=4$ cm, $A D=5$ cm, $D E=1$ cm, $A C=12$ cm, $A E=6$ cm. If the area of triangle $E A B$ is 24 square cm. Then the distance from point $A$ to $C D$ is $\frac{4}{13}$ - .
【Analysis】From $A B=8 \text{ cm}, B C=4 \text{ cm}, A C=12 \text{ cm}$, we know that points $A, B, C$ cannot form a triangle, so points $A, B, C$ are collinear. Similarly, points $A, D, E$ are also collinear. Therefore, we can draw the diagram. Given $A E=6 \text{ cm}, A B=8 \text{ cm}, \triangle E A B$ has an area of $24 \text{ cm}^2$, we can determine that triangle $A B E$ is a right triangle, i.e., $\angle E A B=90^{\circ}$. Thus, triangle $A D C$ is also a right triangle. Using the Pythagorean theorem, it is not difficult to find the distance from point $A$ to $C D$.
【Solution】According to the analysis, points $A, B, C$ are collinear, and points $A, D, E$ are also collinear. The diagram is as follows:
$$
S_{\triangle \mathrm{EAB}}=\frac{1}{2} \times \mathrm{AE} \times \mathrm{AB} \cdot \sin \angle \mathrm{EAF}=24 \text{, i.e., } \frac{1}{2} \times 6 \times 8 \times \sin \angle E A B=24 \text{, }
$$

So $\sin \angle E A B=1$, thus $\angle E A B=90^{\circ}$. In the right triangle $\triangle A D C$, by the Pythagorean theorem, we have
$$
C D=\sqrt{\mathrm{AC}^{2}+\mathrm{AD}^{2}}=\sqrt{12^{2}+5^{2}}=13(\text{ cm}),
$$

Let $A F \times C D=A D \times A C$, i.e., $A F \times 13=5 \times 12$, so $A F=\frac{60}{13}$, which means the distance from point $A$ to $C D$ is $\frac{60}{13}$. Therefore, the answer is: $\frac{60}{13}$ (or written as $4 \frac{8}{13}$ ).