As shown in the figure, find the sum of the degrees of the 12 angles marked with numbers.

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Geometry, Angles.
We first prove a lemma:
As shown in the figure, line segment $A D$ intersects line segment $B C$ at point $O$, and connect $A B, C D$, then $\angle A+\angle B=\angle C+\angle D$. Proof as follows:
$\because \angle A+\angle B+\angle A O B=180^{\circ}, \angle C+\angle D+\angle C O D=180^{\circ}$ (the sum of the interior angles of a triangle equals $180^{\circ}$);
$\therefore \angle A+\angle B+\angle A O B=\angle C+\angle D+\angle C O D$ (substitution of equal quantities);
$\because \angle A O B=\angle C O D$ (vertical angles are equal);
$\therefore \angle A+\angle B=\angle C+\angle D$ (property of equations).
According to the lemma, we have $\angle 1+\angle 2=\angle 13+\angle 14, \angle 3+\angle 4=\angle 13+\angle 15$;
$$
\begin{array}{l}
\therefore \angle 1+\angle 2+\angle 3+\angle 4=(\angle 13+\angle 14+\angle 15)+\angle 13 ; \\
\because \angle 13+\angle 14+\angle 15=180^{\circ}, \angle 13=\angle 16 ; \\
\therefore \angle 1+\angle 2+\angle 3+\angle 4=180^{\circ}+\angle 16 ;
\end{array}
$$

Similarly, $\angle 5+\angle 6+\angle 7+\angle 8=180^{\circ}+\angle 17, \angle 9+\angle 10+\angle 11+\angle 12=180^{\circ}+\angle 18$;
$$
\begin{array}{l}
\therefore \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8+\angle 9+\angle 10+\angle 11+\angle 12=180^{\circ} \times 3+(\angle 16+\angle 17+\angle 18) ; \\
\because \angle 16+\angle 17+\angle 18=180^{\circ} \\
\therefore \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8+\angle 9+\angle 10+\angle 11+\angle 12=720^{\circ}
\end{array}
$$