In $\triangle A B C$, it is known that $A B=A C$, points $P$ and $Q$ are on sides $A B$ and $A C$ respectively, and $A P=P Q=Q C=\frac{\sqrt{3}}{3} B C$. Find the degree measure of $\angle B A C$.
Connect $C P$.
(1) If $\angle A P C$ is an acute angle, as shown in Figure 1, draw $A E \perp C P$, $A D \perp B C$, with the feet of the perpendiculars at $E$ and $D$ respectively.
From $A P=P Q=Q C$, we get $\angle P A Q=\angle P Q A, \angle Q C P=\angle Q P C$.
Also, $\angle P Q A=\angle Q P C+\angle Q C P$, so $\angle B A C=\angle P A Q=2 \angle Q C P=2 \angle A C E$.
In the isosceles $\triangle A B C$, by the "three lines coincide" property, we have $C D=\frac{1}{2} B C, \angle C A D=\frac{1}{2} \angle B A C$.
Thus, $\angle C A D=\angle A C E$.
Also, $\angle A D C=\angle C E A, A C=C A$, so
$\triangle A D C \cong \triangle C E A$.
Therefore, $A E=C D=\frac{1}{2} B C$.
Thus, in the right $\triangle A P E$,
$$
\sin \angle A P E=\frac{A E}{A P}=\frac{\frac{1}{2} B C}{\frac{\sqrt{3}}{3} B C}=\frac{\sqrt{3}}{2}.
$$

Therefore, $\angle A P E=60^{\circ}$.
Hence, $\angle P A C+\angle A C P$
$$
=\angle B A C+\frac{1}{2} \angle B A C=120^{\circ}.
$$

From ii. $\angle B A C=80^{\circ}$.
(2) If $\angle A P C$ is an obtuse angle, as shown in Figure 2.

Similarly, $\angle A P E=60^{\circ}$.
Hence, $\angle P A C+\angle A C P$
$=\angle B A C+\frac{1}{2} \angle B A C$
$=60^{\circ}$.
Thus, $\angle B A C=40^{\circ}$.
In summary, the measure of $\angle B A C$ is
$80^{\circ}$ or $40^{\circ}$.