Example 7 As shown in Figure 6, in the regular nonagon ABCDEFGHI, it is known that $A E=1$. Then the length of $A B + A C$ is



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Solving: Since the sum of the interior angles of a regular nonagon is
$$
(9-2) \times 180^{\circ}=1260^{\circ} \text {, }
$$

and each interior angle is $140^{\circ}$, therefore,
$$
\angle C A B=\left(180^{\circ}-140^{\circ}\right) \div 2=20^{\circ} \text {. }
$$

Connecting $A H$, and drawing $H M$ and $G N$ perpendicular to $A E$ at points $M$ and $N$ respectively.
Then $\angle H A M=3 \times 20^{\circ}=60^{\circ}$
$$
\Rightarrow \angle A H M=30^{\circ}, A H=2 A M \text {. }
$$

Let $A M=E N=x, M N=y$.
Since quadrilateral $H G N M$ is a rectangle, $H G=y$, which is the side length of the regular nonagon.
Thus, $A B+A C=y+2 x$.
And $A E=x+y+x=1$, so $A B+A C=1$.