Example 9 As shown in Figure 19-11, the vertex $A$ of parallelogram $ABCD$ is on the edge $MN$ of the dihedral angle $\alpha-MN-\beta$, and points $B, C, D$ are all on $\alpha$, with $AB=2AD, \angle DAN=45^{\circ}, \angle BAD=60^{\circ}$. Find the cosine value of the plane angle $\varphi$ of the dihedral angle $\alpha-MN-\beta$ so that the projection of parallelogram $ABCD$ on the half-plane $\beta$ is: (I) a rhombus; (II) a rectangle.
From property 2, we know that the projections of $A D$ and $B C$ onto the plane $\beta$ have equal lengths, and the projections of $A B$ and $C D$ onto $\beta$ also have equal lengths. Therefore, the projection of the parallelogram $A B C D$ onto $\beta$ is also a parallelogram (or a line segment).
From property 8, we know that the area of the projection is $S_{A B C D} \cdot \cos \varphi$.
Let the angle between $A D$ and the plane $\beta$ be $x$. Then, from property 5, we have
\[
\frac{A D}{\sin \varphi} = \frac{A D \cdot \sin 45^{\circ}}{\sin x}, \quad \text{i.e.,} \quad \sin x = \frac{\sqrt{2}}{2} \sin \varphi.
\]

Here, we use the fact that the angle between $A D$ and $M N$ is $45^{\circ}$ (thus the distance from $D$ to $M N$ is $A D \cdot \sin 45^{\circ}$). Therefore, if the projections of $B, C, D$ onto the plane $\beta$ are $B^{\prime}, C^{\prime}, D^{\prime}$, respectively, then we have $A D^{\prime} = \sqrt{1 - \frac{1}{2} \sin^2 \theta} \cdot A D$.
Similarly, $A B^{\prime} = \sqrt{1 - \sin^2 75^{\circ} \cdot \sin^2 \varphi} \cdot A B$.
(I) If $A B^{\prime} C^{\prime} D^{\prime}$ is a rhombus, then $A B^{\prime} = A D^{\prime}$, i.e.,
\[
\sqrt{1 - \frac{1}{2} \sin^2 \varphi} \cdot A B = \sqrt{1 - \sin^2 75^{\circ} \cdot \sin^2 \varphi} \cdot A B.
\]

Thus, $1 - \frac{1}{2} \sin^2 \varphi = 4 \left(1 - \sin^2 75^{\circ} \cdot \sin^2 \varphi\right) = 4 - 2 \left(1 - \cos 150^{\circ}\right) \cdot \sin^2 \varphi$.
Therefore, $\sin^2 \theta = 4 \sqrt{3} - 6$, and hence $\cos \theta = 2 - \sqrt{3}$ is the desired result.
(II) If $A B^{\prime} C^{\prime} D^{\prime}$ is a rectangle, then $S_{A B C D} = A B^{\prime} \cdot A D^{\prime}$. Thus, $S_{A B C D} \cdot \cos \varphi = A B^{\prime} \cdot A D^{\prime}$, i.e.,
\[
\frac{\sqrt{3}}{2} \cos \varphi \cdot A B \cdot A D = \sqrt{1 - \frac{1}{2} \sin^2 \varphi} = \sqrt{1 - \sin^2 75^{\circ} \cdot \sin^2 \varphi} \cdot A B \cdot A D.
\]

Therefore, $\frac{3}{4} \cos^2 \varphi = \left(1 - \frac{1}{2} \sin^2 \theta\right) \cdot \left(1 - \frac{1 + \frac{\sqrt{3}}{2}}{2} \sin^2 \varphi\right)$.
Thus, let $t = \sin^2 \varphi$, then $(2 + \sqrt{3}) t^2 - 2(1 + \sqrt{3}) t + 2 = 0$,
i.e., $t = \frac{2(1 + \sqrt{3})}{2(2 + \sqrt{3})} = \sqrt{3} - 1$, and hence $\cos \theta = \sqrt{2 - \sqrt{3}} = \frac{1}{2} (\sqrt{6} - \sqrt{2})$ is the desired result.