Example 2 In $\triangle A B C$, it is known that $\angle A: \angle B: \angle C = 4: 2: 1, \angle A, \angle B, \angle C$ are opposite to sides $a, b, c$ respectively.
(1) Prove: $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$;
(2) Find the value of $\frac{(a+b-c)^{2}}{a^{2}+b^{2}+c^{2}}$.
(1) Proof As shown in Figure 3, construct the angle bisector of $\angle ABC$, intersecting the circumcircle of $\triangle ABC$ at point $M$, and construct the angle bisector of $\angle BAC$, intersecting the circumcircle of $\triangle ABC$ at point $N$.
Then $AM=MC=AB=c, AM \parallel BC$;
$$
CN=NB=AC=b, AB \parallel CN \text{. }
$$

Thus, quadrilaterals $ABCM$ and $ABNC$ are both isosceles trapezoids, and we have
$$
BM=AC=b, AN=BC=a.
$$

Note that $M$ and $N$ are the midpoints of arcs $\overparen{AC}$ and $\overparen{BC}$ (excluding $B$ or $A$), respectively. By Corollary 2(2), we have
$$
\begin{array}{l}
MB^2 - MA^2 = AB \cdot BC, \\
NA^2 - NB^2 = AB \cdot AC,
\end{array}
$$

which gives $b^2 - c^2 = ac, a^2 - b^2 = bc$.
Adding these two equations, we get
$$
\begin{array}{l}
a^2 - ac - c(b+c) \\
\Rightarrow \frac{b+c}{a} = \frac{a-c}{c} = \frac{a}{c} - 1.
\end{array}
$$

Also, from $b^2 + bc = a^2$, we have
$$
\frac{b+c}{a} = \frac{a}{b}.
$$

From equations (1) and (2), we get
$$
\frac{a}{c} - \frac{a}{b} = 1 \Rightarrow \frac{1}{a} + \frac{1}{b} = \frac{1}{c}.
$$
(2) Solution From equation (3), we have $ab = ac + bc$.

Thus, $\frac{(a+b-c)^2}{a^2 + b^2 + c^2}$
$$
= \frac{a^2 + b^2 + c^2 + 2(ab - bc - ac)}{a^2 + b^2 + c^2} = 1.
$$