Firstly, for each side of the triangle, we draw 37 equidistant, parallel lines to it. In this way we get 382 = 1444 triangles. Then we erase 11 lines which are closest to the vertex A and parallel to the side BC and we draw 21 lines perpendicular to BC, 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 · 21 + 10 = 556 new triangles. Therefore we obtain a total of 2000 triangles and we have used 37 · 3 − 11 + 21 = 121 lines. Let D be the 12th point on side AB, starting from B (including it). The perpendicular to BC passing through D will be the last line we draw. In this way we obtain the required configuration.