Example 3 㷵 On a board, there is a convex 2011-gon, and Betya draws its diagonals one by one. It is known that each diagonal drawn intersects at most one of the previously drawn diagonals at an interior point. Question: What is the maximum number of diagonals Betya can draw? [3]
To prove by induction: For a convex $n$-sided polygon, at most $2n-6$ diagonals can be drawn.

Let $A_{1} A_{2} \cdots A_{n}$ be a convex polygon. The following $2n-6$ diagonals can be drawn sequentially:
$A_{2} A_{4}, A_{3} A_{5}, \cdots, A_{n-2} A_{n}, A_{1} A_{3}, A_{1} A_{4}, \cdots, A_{1} A_{n-1}$.
When $n=3$, the conclusion is obviously true.
Assume the proposition holds for all values less than $n$.
For a convex $n$-sided polygon, let the last diagonal drawn be $A_{1} A_{k}$, which intersects at most one of the previously drawn diagonals (if it exists, let it be $l$) at an interior point. All the diagonals drawn, except for $A_{1} A_{k}$ and $l$, lie entirely within the $k$-sided polygon $A_{1} A_{2} \cdots A_{k}$ or the $(n+2-k)$-sided polygon $A_{k} A_{k+1} \cdots A_{n} A_{1}$.
By the induction hypothesis, there are at most
$$
(2k-6) + [2(n+2-k)-6] = 2n-8
$$
diagonals; adding $A_{1} A_{k}$ and $l$ results in at most $2n-6$ diagonals.
Therefore, for a convex $n$-sided polygon, at most $2n-6$ diagonals can be drawn.
Thus, Betja can draw at most 4016 diagonals.