Using diagonals from a common vertex, how many triangles could be formed from a nonagon?

Let the vertices of the nonagon be labeled \( V_1, V_2, \dots, V_9 \). Choose a common vertex, say \( V_1 \). From \( V_1 \), you can draw diagonals to all the vertices that are not adjacent to \( V_1 \) (i.e. not \( V_2 \) or \( V_9 \)). This gives: \[ 9 - 3 = 6 \quad \text{diagonals.} \] Drawing these diagonals divides the nonagon into triangles. The total number of triangles formed is the same as the number of regions the polygon is divided into, which for a convex \( n \)-gon (with \( n=9 \)) by drawing all diagonals from one vertex is: \[ n - 2 = 9 - 2 = 7. \] So, the triangles formed are: - Triangle \( V_1V_2V_3 \) - Triangle \( V_1V_3V_4 \) - Triangle \( V_1V_4V_5 \) - Triangle \( V_1V_5V_6 \) - Triangle \( V_1V_6V_7 \) - Triangle \( V_1V_7V_8 \) - Triangle \( V_1V_8V_9 \) Thus, the number of triangles formed is \( \boxed{7} \).