7.23 Through a non-central point inside a regular 12-sided polygon, what is the maximum number of different diagonals that can be drawn?
[Solution] In the right figure, point $M$ is both the incenter of $\triangle A_{5} A_{11} A_{3}$ and the incenter of $\triangle A_{2} A_{4} A_{8}$, so the four diagonals $A_{1} A_{5}, A_{2} A_{6}, A_{3} A_{8}$, $A_{4} A_{11}$ intersect at point $M$. This indicates that the maximum value sought is no less than 4.

For convenience, we use arc length to represent the length of the chord subtended by the arc, and when the arc length is $\frac{k}{12}$ of the circumference, we say the chord length is $k, k \leqslant 6$. If 5 diagonals intersect at a non-center point inside the polygon, then each chord length is no less than 5, and at most one of them has a length of 6. Because a chord of length 6 is a diameter, if there are two diagonals of length 6 intersecting at a point, the intersection point is the center. This means that among the 5 diagonals intersecting at a point, at least 4 have a length of 5. To prove that it is impossible for 5 diagonals to intersect at a non-center point inside the polygon, it is only necessary to prove the following lemma.

Lemma: Any 3 diagonals of length 5 cannot intersect at the same point.

Proof of the Lemma: Clearly, the intersection of two chords of length 5 can only have the following 4 different cases: $\left\{A_{1} A_{6}, A_{2} A_{9}\right\}$, $\left\{A_{1} A_{6}, A_{2} A_{7}\right\},\left\{A_{1} A_{6}, A_{3} A_{10}\right\},\left\{A_{1} A_{6}\right.$, $\left.A_{3} A_{8}\right\}$, let their intersection points be $M_{1}, M_{2}, M_{3}$, $M_{4}$, then these 4 points are neither the same nor symmetric about the midpoint of the segment $A_{1} A_{6}$.

In fact, by the Law of Sines, $A_{1} M_{1}: M_{1} A_{6}=A_{1} M_{1}: M_{1} A_{9}=\sin A_{9} : \sin A_{1}=\sin \alpha: \sin 3 \alpha$, where $\alpha=15^{\circ}$. Similarly, $A_{1} M_{2}: M_{2} A_{6}=\sin 6 \alpha: \sin 4 \alpha, A_{1} M_{3}: M_{3} A_{6}=\sin 2 \alpha: \sin 4 \alpha, A_{1} M_{4}: M_{4} A_{6}=\sin 5 \alpha: \sin 3 \alpha$. It is not difficult to verify that none of these 4 ratios are equal to each other, nor is any one the reciprocal of another. This proves the lemma.
In summary, the maximum value sought in the problem is 4.