3. In $\triangle A B C$, it is known that $A B=A C, D$ is a point on $A C$, and $A D=B D=B C$. Then $\cos \angle B A C=(\quad)$.
(A) $\frac{\sqrt{5}-1}{2}$
(B) $\frac{\sqrt{5}+1}{2}$
(C) $\frac{\sqrt{5}+1}{4}$
(D) $\frac{\sqrt{5}-1}{4}$
3. C.

As shown in Figure 3, let $A B=A C=1$, $A D=B D=B C=a$. Then
$$
C D=1-a \text {. }
$$

It is easy to see that,
$$
\begin{array}{l}
\triangle A B C \backsim \triangle B C D \\
\Rightarrow \frac{A B}{B C}=\frac{B C}{C D} \Rightarrow \frac{1}{a}=\frac{a}{1-a} \\
\Rightarrow a^{2}=1-a \\
\Rightarrow a=\frac{\sqrt{5}-1}{2} \text { (negative value discarded). }
\end{array}
$$

Draw $B H \perp C D$ at point $H$. Then
$$
C H=D H=\frac{1}{2}(1-a)=\frac{3-\sqrt{5}}{4} \text {. }
$$

Also, $A H=1-C H=1-\frac{3-\sqrt{5}}{4}=\frac{\sqrt{5}+1}{4}$, thus
$$
\cos \angle B A C=\frac{A H}{A B}=\frac{\sqrt{5}+1}{4} .
$$