Example 16 (2003 National College Entrance Examination Question) As shown in Figure 7-15, in the right triangular prism $A B C-A_{1} B_{1} C_{1}$, the base is an isosceles right triangle, $\angle A C B=90^{\circ}$, the side edge $A A_{1}=2$, $D$ and $E$ are the midpoints of $C_{1} C$ and $A_{1} B$ respectively, and the projection of point $E$ on plane $A B D$ is the centroid $G$ of $\triangle A B D$.
(1) Find the size of the angle formed by $A_{1} B$ and plane $A B D$ (express the result using the value of an inverse trigonometric function);
(2) Find the distance from point $A_{1}$ to plane $A E D$.
(1) Connect $B G$, then $B G$ is the projection of $B E$ on the plane $A B D$, i.e., $\angle E B G$ is the angle formed by $A_{1} B$ and the plane $A B D$. Let $F$ be the midpoint of $A B$, and connect $E F, F C$. Since $D, E$ are the midpoints of $C C_{1}, A_{1} B$ respectively, and $D C \perp$ plane $A B C$, then $C D E F$ is a rectangle.

Connect $D F, G$ is the centroid of $\triangle A D B$, then $G$ lies on $D F$. In the right triangle $\triangle E F D$, $E F^{2}=F G \cdot F D=$ $\frac{1}{3} F D^{2}$. Since $E F=1$, then $F D=\sqrt{3}$.
Thus, $E D=\sqrt{2}, E G=\frac{1 \cdot \sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3}$.
Also, $F C=E D=\sqrt{2}$, then $A B=2 \sqrt{2}, A_{1} B=2 \sqrt{3}, E B=\sqrt{3}$.
Therefore, $\sin \angle E B G=\frac{E G}{E B}=\frac{\sqrt{6}}{3} \cdot \frac{1}{\sqrt{3}}=\frac{\sqrt{2}}{3}$, i.e., the angle formed by $A_{1} B$ and the plane $A B D$ is $\arcsin \frac{\sqrt{2}}{3}$.
(2) Since $E D \perp A B, E D \perp E F$, and $E F \cap A B=F$, then $E D \perp$ plane $A_{1} A B$.

Since $E D \subset$ plane $A E D$, then plane $A E D \perp$ plane $A_{1} A B$, and plane $A E D$ $\cap$ plane $A_{1} A B=A E$. Draw $A_{1} K \perp A E$ at $K$, then $A_{1} K \perp$ plane $A E D$, i.e., $A_{1} K$ is the distance from $A_{1}$ to plane $A E D$. In $\triangle A_{1} A B$, $A_{1} K=$ $\frac{A_{1} A \cdot A_{1} B_{1}}{A B_{1}}=\frac{2 \cdot 2 \sqrt{2}}{2 \sqrt{3}}=\frac{2 \sqrt{6}}{3}$, hence the distance from $A_{1}$ to plane $A E D$ is $\frac{2 \sqrt{6}}{3}$.
Alternative solution Connect $A_{1} D$, then $V_{A_{1}-A D E}=V_{D-A A_{1} E}$.
Since $E D \perp A B, E D \perp E F$, and $E F \cap A B=F$, then $E D \perp$ plane $A_{1} A B$.
Let the distance from $A_{1}$ to plane $A E D$ be $h$, then $S_{\triangle A E D} \cdot h=S_{\triangle A_{1} A E} \cdot E D$.
Also, $S_{\triangle A_{1} A E}=\frac{1}{2} S_{\triangle A_{1} A B}=\frac{1}{4} A_{1} A \cdot A B=\sqrt{2}, S_{\triangle A E D}=\frac{1}{2} A E \cdot E D=\frac{\sqrt{6}}{2}$, thus $h=$ $\sqrt{2} \cdot \sqrt{2} / \frac{\sqrt{6}}{2}=\frac{2}{3} \sqrt{6}$, i.e., the distance from $A_{1}$ to plane $A E D$ is $\frac{2 \sqrt{6}}{3}$.