Three, it is known that $A D$ is the altitude from vertex $A$ to side $B C$ of $\triangle A B C$, and $B C + A D = A B + A C$. Find the range of values for $\angle A$.

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Three, let $A B=c, B C=a, C A=b, A D=h$. From the triangle area formula, we have $b c \sin A=a h$, then
$$
b c=\frac{a h}{\sin A} \text {. }
$$

From $B C+A D=A B+A C$, we get
$$
b+c=a+h .
$$

From the cosine rule, we have
$$
\begin{array}{l}
\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{(b+c)^{2}-a^{2}-2 b c}{2 b c} \\
=\frac{(a+h)^{2}-a^{2}}{\frac{2 a h}{\sin A}}-1=\left(1+\frac{h}{2 a}\right) \sin A-1 \text {. } \\
\end{array}
$$

Therefore, $\sin A=\frac{1+\cos A}{1+\frac{h}{2 a}}=\frac{2 \cos ^{2} \frac{A}{2}}{1+\frac{h}{2 a}}$, which means
$$
\cot \frac{A}{2}=1+\frac{h}{2 a} \text {. }
$$

As shown in Figure 2, construct $C E \perp B C$, such that $C E=2 h$, and connect $B E$ and $A E$.
In the right triangle $\triangle B C E$, we have
$$
B E=\sqrt{a^{2}+4 h^{2}} \text {, }
$$

and $a+h=b+c \geqslant B E$.
Thus, $a+h$
$$
\geqslant \sqrt{a^{2}+4 h^{2}} \text {. }
$$

Solving this, we get $\frac{h}{a} \leqslant \frac{2}{3}$. Hence, $1+\frac{h}{2 a} \in\left(1, \frac{4}{3}\right]$.
Therefore, $\cot \frac{A}{2} \in\left(1, \frac{4}{3}\right]$.
Thus, $\angle A \in\left[2 \operatorname{arccot} \frac{4}{3}, \frac{\pi}{2}\right)$.