We are given a triangle $ABC$ with points on its sides, and we are tasked with finding how many triangles can be formed such that:

• One vertex is on side $AB$,
• One vertex is on side $BC$,
• The third vertex is anywhere on side $AC$ (with no restriction).

The points on each side are distinct and do not coincide with the vertices of the triangle. The number of points on each side is as follows:

• 6 points on side $AB$,
• 7 points on side $BC$,
• 8 points on side $AC$.

Step 1: Understand the problem

To form a triangle, we need to:

1. Choose one point from side $AB$,
2. Choose one point from side $BC$,
3. Choose one point from side $AC$.

There are no restrictions on the points from side $AC$, so we can choose any of the 8 points from this side.

Step 2: Count the number of triangles

The number of triangles is determined by the number of ways to choose:

• 1 point from the 6 points on side $AB$,
• 1 point from the 7 points on side $BC$,
• 1 point from the 8 points on side $AC$.

The total number of triangles is the product of the number of choices from each side:

$\text{Number of triangles} = 6 \times 7 \times 8 = 336$

Final Answer:

There are $336$ triangles that can be formed with one vertex on $AB$, one vertex on $BC$, and one vertex anywhere on $AC$.

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Do you have any questions or would you like further clarification?

Here are 5 related questions for further exploration:

1. What if there were 9 points on side $AC$? How would this change the total number of triangles?
2. How many triangles can be formed if two vertices must lie on the same side of the triangle?
3. What happens if one of the sides, such as $AC$, only has 5 points? How many triangles can be formed then?
4. How would the number of triangles change