1. In tetrahedron $ABCD$, $AD \perp$ plane $BCD$, $\angle ABD = \angle BDC = \theta < 45^{\circ}$. It is known that $E$ is a point on $BD$ such that $CE \perp BD$, and $BE = AD = 1$.
(1) Prove: $\angle BAC = \theta$;
(2) If the distance from point $D$ to plane $ABC$ is $\frac{4}{13}$, find the value of $\cos \theta$.


1. From $A D=B E=1$, we have
$$
\begin{array}{l}
A B=\frac{1}{\sin \theta}, B D=\frac{\cos \theta}{\sin \theta}, D E=\frac{\cos \theta}{\sin \theta}-1, \\
C D=\frac{D E}{\cos \theta}=\frac{1}{\sin \theta}-\frac{1}{\cos \theta} .
\end{array}
$$

Thus, $A C=\sqrt{A D^{2}+C D^{2}}$
$$
\begin{array}{c}
=\sqrt{1+\left(\frac{1}{\sin \theta}-\frac{1}{\cos \theta}\right)^{2}} \\
=\frac{1}{\sin \theta \cdot \cos \theta}-1 .
\end{array}
$$

Let $\angle B A C=\alpha$. Using the cosine rule in $\triangle A B C$ and $\triangle D B C$, we get
$$
\begin{array}{l}
B C^{2}=A B^{2}+A C^{2}-2 A B \cdot A C \cos \alpha, \\
B C^{2}=B D^{2}+C D^{2}-2 B D \cdot C D \cos \theta .
\end{array}
$$

Subtracting the above two equations and noting that
$$
A B^{2}=B D^{2}+A D^{2}, A C^{2}=C D^{2}+A D^{2} \text {, }
$$

we get $A D^{2}+B D \cdot C D \cos \theta=A B \cdot A C \cos \alpha$. Therefore, $\cos \alpha=\frac{A D^{2}+B D \cdot C D \cos \theta}{A B \cdot A C}$
$$
=\frac{1+\frac{\cos \theta}{\sin \theta}\left(\frac{1}{\sin \theta}-\frac{1}{\cos \theta}\right) \cos \theta}{\frac{1}{\sin \theta}\left(\frac{1}{\sin \theta \cdot \cos \theta}-1\right)}=\cos \theta \text {. }
$$

Thus, $\angle B A C=\theta$.
Noting that the volume of the tetrahedron $A B C D$ is
$$
\begin{array}{l}
V=\frac{1}{6} A D \cdot B D \cdot C D \sin \theta, \\
S_{\triangle A B C}=\frac{1}{2} A B \cdot A C \sin \theta .
\end{array}
$$

Therefore, the distance from point $D$ to the plane $A B C$ is
$$
\begin{array}{l}
\frac{3 V}{S_{\triangle A B C}}=\frac{A D \cdot B D \cdot C D}{A B \cdot A C} \\
=\frac{\cos ^{2} \theta-\cos \theta \cdot \sin \theta}{1-\sin \theta \cdot \cos \theta}=\frac{4}{13} .
\end{array}
$$

Solving this, we get $\cos \theta=\frac{4}{5}$.