(13) In the 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 $\cos \theta$.
13 (1) Since $A D=B E=1$, we have $A B=\frac{1}{\sin \theta}, B D=\frac{\cos \theta}{\sin \theta}, D E=$ $\frac{\cos \theta}{\sin \theta}-1$, and
$$
C D=\frac{D E}{\cos \theta}=\frac{1}{\sin \theta}-\frac{1}{\cos \theta} .
$$

Thus, we get
$$
A C=\sqrt{A D^{2}+C D^{2}}=\sqrt{1+\left(\frac{1}{\sin \theta}-\frac{1}{\cos \theta}\right)^{2}}=\frac{1}{\sin \theta \cos \theta}-1 .
$$

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

Subtracting the above two equations and noting that
$$
\begin{array}{l}
A B^{2}=B D^{2}+A D^{2} \\
A C^{2}=C D^{2}+A D^{2}
\end{array}
$$

we get
$$
2 A D^{2}-2 A B \cdot A C \cdot \cos \alpha+2 B D \cdot C D \cdot \cos \theta=0,
$$

thus
$$
\begin{aligned}
\cos \alpha & =\frac{A D^{2}+B D \cdot C D \cdot \cos \theta}{A B \cdot A C} \\
& =\frac{1+\frac{\cos \theta}{\sin \theta} \cdot\left(\frac{1}{\sin \theta}-\frac{1}{\cos \theta}\right) \cdot \cos \theta}{\frac{1}{\sin \theta} \cdot\left(\frac{1}{\sin \theta \cos \theta}-1\right)} \\
& =\cos \theta .
\end{aligned}
$$

This proves that $\angle B A C=\theta$.
(2) Note that the volume of the tetrahedron $A B C D$ is
$$
V=\frac{1}{3} A D \cdot S_{B C D}=\frac{1}{6} A D \cdot B D \cdot C D \cdot \sin \theta,
$$

and the area of $\triangle A B C$ is
$$
S=\frac{1}{2} A B \cdot A C \cdot \sin \theta,
$$

Therefore, the distance from point $D$ to the plane $A B C$ is
$$
\frac{3 V}{S}=\frac{A D \cdot B D \cdot C D}{A B \cdot A C}=\frac{\cos ^{2} \theta-\cos \theta \sin \theta}{1-\sin \theta \cos \theta} .
$$

Setting the above expression equal to $\frac{4}{13}$, we solve to get $\cos \theta=\frac{4}{5}$.