Example 1. As shown in Figure 1, the edge length of the cube is $\mathrm{a}$. (1) Find the angle $\theta$ between $\mathrm{AC}^{\prime}$ and $\mathbf{D B}^{\prime}$, (2) Draw a perpendicular line $\mathbf{A E}$ from $\mathbf{A}$ to the diagonal $\mathbf{A}^{\prime} \mathrm{C}$, prove that $\frac{\mathrm{AE}^{2}}{\mathrm{EC}^{2}}=\frac{1}{2}$.
(1) As long as we guide the students to observe that $\mathrm{AC}^{\prime}$ and $\mathrm{DB}^{\prime}$ lie on the diagonal plane $\mathrm{DAB}^{\prime} \mathrm{C}^{\prime}$, the problem clearly becomes one of finding the angle between the two diagonals of the rectangle $\mathrm{SAR}^{\prime \prime} \mathrm{C}^{\prime}$ (see Figure 2).
By the cosine rule,
$$
\begin{array}{l}
\cos \theta=\frac{\mathrm{OC}^{\prime 2}+\mathrm{OB}^{\prime 2}-\mathrm{C}^{\prime} \mathrm{B}^{2}}{2 \mathrm{OC}^{\prime} \cdot \mathrm{OB}^{\prime}} \\
=\frac{2\left(\frac{\sqrt{ } 3}{2} \mathrm{a}\right)^{2}-\mathrm{a}^{2}}{2\left(\frac{\sqrt{3}}{2} \mathrm{a}\right)^{2}}=\frac{1}{3} . \\
\text { Hence } \theta=\arccos \frac{1}{3} .
\end{array}
$$
(2) Guide the students to observe the plane determined by the two perpendicular intersecting line segments $\mathrm{AE}$ and $\mathbf{A}^{\prime} \mathrm{C}^{\prime}$, which is exactly the plane where the triangle $\mathrm{AA}^{\prime} \mathrm{C}$ lies. Connect $\mathbf{A C}$, then the problem is transformed into finding the relationship in the right-angled $\triangle \mathbf{A}^{\prime} \mathbf{A C}$, where $\frac{\mathrm{AE}^{2}}{\mathrm{EC}^{2}}=\frac{1}{2}$ holds (as shown in Figure 3).
$$
\begin{array}{l}
\because \triangle \mathbf{A}^{\prime} \mathrm{AE} \sim \triangle \mathrm{ACE}, \\
\therefore \frac{\mathrm{AE}}{\mathrm{EC}}=\frac{A A^{\prime}}{\mathrm{AC}}=\frac{a}{\sqrt{2} a}=\frac{\sqrt{2}}{2}, \\
\therefore \frac{\mathrm{AE}^{2}}{\mathrm{EC}^{2}}=\frac{1}{2} .
\end{array}
$$