II. (25 points) In $\triangle A B C$, $A D$ is the angle bisector, and $\frac{1}{A D}=\frac{1}{A B}+\frac{1}{A C}$. Find the measure of $\angle B A C$.
As shown in Figure 5, draw $B E \parallel A D$ intersecting the extension of $C A$ at $E$.
From $\frac{1}{A D}=\frac{1}{A B}+\frac{1}{A C}$
$=\frac{A B+A C}{A B \cdot A C}$, we get
$A B+A C=\frac{A B \cdot A C}{A D}$.
(1)
From $\frac{B D}{D C}=\frac{A B}{A C}$, we get
$\frac{B C}{D C}=\frac{A B + A C}{A C}$.
From (1) and (2), we get
$\frac{B C}{D C}=\frac{A B \cdot A C}{A D \cdot A C}=\frac{A B}{A D}$.
Since $\frac{B C}{D C}=\frac{B E}{A D}$,
it follows that $B E=A B$.
Given $A D \parallel B E$, we know $\angle 1=\angle 2=\angle 3=\angle E$,
thus $A E=A B$.
Therefore, $\triangle A B E$ is an equilateral triangle, $\angle B A E=60^{\circ}$.
Hence, $\angle B A C=120^{\circ}$.