9. As shown in the figure, in the right dihedral angle $D-A B-E$, quadrilateral $A B C D$ is a square with a side length of 2, $A E=E B$, and $F$ is a point on $C E$ such that $B F \perp$ plane $A C E$.
(1) Find the size of the dihedral angle $B-A C-E$;
(2) Find the distance from point $D$ to the plane $A C E$.
9. Solution 1 (1) Since $B F \perp$ plane $A C E$. Therefore, $B F \perp A E$.

Since the dihedral angle $D-A B-E$ is a right dihedral angle, and $C B \perp A B$,
Therefore, $C B \perp$ plane $A B E$, so $C B \perp A E$.
Therefore, $A E \perp$ plane $B C E$, so $A E \perp E B$,
Also, $A E=E B$, hence $\triangle A E B$ is an isosceles right triangle, so $A E=B E=\sqrt{2}$.
Connecting $B D$ intersects $A C$ at $G$, connecting $F G$,
Since the side length of the square $A B C D$ is 2, so $B G \perp A C, B G=\sqrt{2}$,
Since $B F \perp$ plane $A C E$,
By the inverse of the three perpendiculars theorem, we get $F G \perp A C$. Therefore, $\angle B G F$ is the plane angle of the dihedral angle $B-A C-E$.
In the right $\triangle B C E$, $E C=\sqrt{B C^{2}+B E^{2}}=\sqrt{6}, B F=\frac{B C \cdot B E}{E C}=\frac{2 \times \sqrt{2}}{\sqrt{6}}=\frac{2 \sqrt{3}}{3}$,
Therefore, in the right $\triangle B F G$, $\sin \angle B G F=\frac{B F}{B G}=\frac{\frac{2 \sqrt{3}}{3}}{\sqrt{2}}=\frac{\sqrt{6}}{3}$.
Therefore, the dihedral angle $B-A C-E$ equals $\arcsin \frac{\sqrt{6}}{3}$.
(2) Draw $E O \perp A B$ at point $O, E O=1$.
Since the dihedral angle $D-A B-E$ is a right dihedral angle, so $E O \perp$ plane $A B C D$.
Let the distance from $D$ to plane $A C E$ be $h$,
Since $V_{D-A C E}=V_{E-A C D}$, so $\frac{1}{3} S_{\triangle A C E} \cdot h=\frac{1}{3} S_{\triangle A C D} \cdot E O$.
Since $A E \perp$ plane $B C E$, so $A E \perp E C$.
Therefore, $h=\frac{\frac{1}{2} A D \cdot D C \cdot E O}{\frac{1}{2} A E \cdot E C}=\frac{\frac{1}{2} \times 2 \times 2 \times 1}{\frac{1}{2} \times \sqrt{2} \times \sqrt{6}}=\frac{2 \sqrt{3}}{3}$.
Therefore, the distance from point $D$ to plane $A C E$ is $\frac{2 \sqrt{3}}{3}$.
Solution 2 (1) As shown in the figure, take the midpoint of line segment $A B$ as the origin $O, O E$ as the $x$-axis, $A B$ as the $y$-axis, and establish a spatial rectangular coordinate system $O-x y z$.
Similarly to solution 1, $A E \perp$ plane $B C E, B E \subset$ plane $B C E$, so $A E \perp B E$,
In the right $\triangle A E B$, $A B=2, O$ is the midpoint of $A B$, so $O E=1$,
Therefore, $A(0,-1,0), E(1,0,0), C(0,1,2)$.
$$
\overrightarrow{A E}=(1,1,0), \overrightarrow{A C}=(0,2,2) \text {. }
$$

Let a normal vector of plane $A E C$ be $\boldsymbol{n}=(x, y, z)$,
Then $\left\{\begin{array}{l}\overrightarrow{A E} \cdot n=0, \\ \overrightarrow{A C} \cdot n=0,\end{array}\right.$ which is $\left\{\begin{array}{l}x+y=0, \\ 2 y+2 z=0 .\end{array}\right.$
Let $x=1$, we get $n=(1,-1,1)$ is a normal vector of plane $A E C$.
Also, a normal vector of plane $B A C$ is $\boldsymbol{m}=(1,0,0)$,
Therefore, $\cos \langle\boldsymbol{m}, \boldsymbol{n}\rangle=\frac{\boldsymbol{m} \cdot \boldsymbol{n}}{|\boldsymbol{m}| \cdot|\boldsymbol{n}|}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$.
Therefore, the size of the dihedral angle $B-A C-E$ is $\arccos \frac{\sqrt{3}}{3}$.
(2) Since $A D / / z$-axis, $A D=2$, so $\overrightarrow{A D}=(0,0,2)$,

Therefore, the distance from point $D$ to plane $A C E$ is $d=|\overrightarrow{A D}| \cdot|\cos \langle\overrightarrow{A D}, \boldsymbol{n}\rangle|=\frac{|\overrightarrow{A D} \cdot n|}{|n|}=\frac{2}{\sqrt{3}}=\frac{2}{3} \sqrt{3}$.
10. Solution Let $L M$ be the common perpendicular segment of lines $l$ and $m$, $L \in l, M \in m$.
Draw a line $l^{\prime} / / l$ through $M$, and the plane determined by $l^{\prime}$ and $m$ is denoted as plane $P$, obviously, line $L M \perp$ plane $P$.
Draw lines $A G, B H, C K$ parallel to line $L M$ through points $A, B, C$ respectively, intersecting line $l^{\prime}$ at $G, H, K$, since $A B=B C$, $A G / / H B / / C K$, hence $G H=H K$.
Also, since $A D \perp m, B E \perp m, C F \perp m$, by the three perpendiculars theorem, $G D \perp m, H E \perp m, K F \perp m$,
Therefore, $G D / / H E / / K F$, and $E, H$ are the midpoints of $F D, K G$ respectively.