It is well-known problem, but quite now I am unable to find a link on AoPS. For any line $a$ take all $n-1$ points, in which it meets other lines, and for any two consecutive points $B=a\cap b$, $C=a\cap c$ consider the triangle, formed by lines $a$, $b$, $c$ and draw a flower inside this triangle near the midpoint of its side $BC$. Totally, we get $(n-2)n$ flowers. On the other hand, in any part, which is not a triangle, we have at most two flowers (because any two flowers in the same part must lie on neighbouring sides of this part). Since we have $n(n+1)/2+1$ parts (simple induction), and $2n$ of them are unbounded (common sense), we get at most $3T+2(n(n+1)/2+1-2n-T)$ flowers, hence $(n-2)n\leq n^2-3n+2+T$, $T\geq n-2$, where $T$ is the number of triangular parts.