9.3. There is a convex 2011-gon on the blackboard. 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?
9.3.4016.

Use induction to prove: 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. We can sequentially draw $2n-6$ diagonals as follows: $A_{2} A_{4}, A_{3} A_{5}$, $A_{4} A_{6}, \cdots, A_{n-2} A_{n}, A_{1} A_{3}, A_{1} A_{4}, \cdots, A_{1} A_{n-1}$.

We will use mathematical induction to prove that at most $2n-6$ diagonals can be drawn.

When $n=3$, the conclusion is obvious. Assume that the last diagonal drawn is $A_{1} A_{k}$. It can intersect at most one of the previously drawn diagonals (if it exists, let it be $d$) at an internal point. All the drawn diagonals, except for $A_{1} A_{k}$ and $d$, are entirely within the $k$-sided polygon $A_{1} A_{2} \cdots A_{k}$ and the $(n+2-k)$-sided polygon $A_{k} A_{k+1} \cdots A_{n} A_{1}$. By the induction hypothesis, we have at most
$$
\begin{array}{l}
(2 k-6)+[2(n+2-k)-6] \\
=2 n-8 \text { (diagonals). }
\end{array}
$$