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

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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 $$ \boxed{7} $$.
From a common vertex in a nonagon, 7 triangles can be formed by drawing diagonals to the non-adjacent vertices.

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

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