As shown in Figure 1, points $D$ and $E$ are on the sides $AB$ and $AC$ of equilateral $\triangle ABC$, respectively, satisfying $BD = AE$. Connecting $CD$ and $BE$ intersect at point $O$. Given that $BO = 2$ and $CO = 5$, find the length of $AO$.

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The above text translated into English, preserving the original text's line breaks and format, is as follows:

As shown in Figure 1, points $D$ and $E$ are on the sides $AB$ and $AC$ of equilateral $\triangle ABC$, respectively, satisfying $BD = AE$. Connecting $CD$ and $BE$ intersect at point $O$. Given that $BO = 2$ and $CO = 5$, find the length of $AO$.
Solving, we know
$$
\begin{array}{l}
\triangle D B C \cong \triangle E A B \\
\Rightarrow \angle B C D=\angle A B E . \\
\text { Then } \angle B O C=180^{\circ}-(\angle B C D+\angle O B C) \\
=180^{\circ}-(\angle A B E+\angle O B C) \\
=180^{\circ}-60^{\circ}=120^{\circ} .
\end{array}
$$

As shown in Figure 2, rotate
$\triangle A B O$ around point $B$
to the position of $\triangle C B P$,
connect $O P$.
Then $\triangle B P O$ is an

equilateral triangle, so
$$
\begin{array}{l}
\angle B O P=60^{\circ}, \\
O P=B O=2 .
\end{array}
$$

Therefore, $\angle P O C=\angle B O C-\angle B O P=60^{\circ}$.
Draw $P F \perp O C$, with the foot of the perpendicular at $F$. Then
$$
\begin{array}{l}
O F=O P \cos 60^{\circ}=1, \\
P F=O P \sin 60^{\circ}=\sqrt{3}, \\
C F=C O-O F=4 .
\end{array}
$$

In the right triangle $\triangle P F C$,
$$
P C=\sqrt{P F^{2}+C F^{2}}=\sqrt{19} \text {. }
$$

Therefore, $A O=P C=\sqrt{19}$.