Firstly, for each side of the triangle, we draw 37 equidistant, parallel lines to it. In this way we get $38^{2}=1444$ triangles. Then we erase 11 lines which are closest to the vertex $A$ and parallel to the side $B C$ and we draw 21 lines perpendicular to $B C$, the first starting from the vertex $A$ and 10 on each of the two sides, the lines which are closest to the vertex $A$, distributed symmetrically. In this way we get $26 \cdot 21+10=$ 556 new triangles. Therefore we obtain a total of 2000 triangles and we have used $37 \cdot 3-11+21=121$ lines. Let $D$ be the $12^{\text {th }}$ point on side $A B$, starting from $B$ (including it). The perpendicular to $B C$ passing through $D$ will be the last line we draw. In this way we obtain the required configuration.