It's easy to count triangles

One important fact to notice is that we cannot cut one of the edges of the triangle with a single line, as this would produce an obtuse angle on one side of the line. In order to create only acute angles, we need another line to intersect at the same point, as shown here.

Let's call this fact (1), as it will be used several times.

Start by assuming that all of the lines that divide up the triangle must intersect on the triangle's edges, and cannot cut each other inside the triangle (this is wrong, as will be demonstrated through a contradiction). Call this assumption (2).

We need to divide the obtuse angle with at least one line, and once that is done we must draw another line from the opposite edge because of (1).

Now, depending on where we placed the second line, we will need to draw a third due to (1) again. However, if we draw this line on the side of the obtuse angle, it will intersect the first line (shown left). Therefore it must be on the other side (shown right).

But, we can see that due to (1) we will need to draw a fourth line, and we end up with the same problem as above. So, whenever we construct a new line, another one must be made to solve (1), and so we can never divide up the triangle in this manner. Hence, (2) is wrong, meaning there must be at least one intersection point in the middle of the triangle.

Let's say $n$ lines intersect at a single point. The only way for all the angles to be acute is if $n\geq5$ .

So, if we place an intersection point in the centre of the triangle, we need at least 5 lines extending from it. One of these lines we can use to cut the obtuse angle in the large triangle, and the other 4 will intersect the sides of the triangle. Due to (1), we will need to draw at least 2 more lines to fix the obtuse angles at these 4 intersection points.

This gives a total of 7 acute triangles . To definitively show that this must be a minimum, we consider every other scenario:

• Adding more lines to the intersection point would not lower the triangle count, as each line will require an additional line to account for (1)
• Adding more intersection points would not lower the triangle count, as each would require 5 lines extending from it, greatly increasing the number of lines.
• Having one line from the intersection cut one of the other corners of the triangle would not lower the triangle count, as we would have 3 lines left to draw, which we cannot join evenly to fix (1) at the end. We could have 6 lines extending from the intersection point, but then adjacent lines would intersect the same sides of the triangle, and so once again we cannot join them.
• Having two lines from the intersection cut both of the other corners of the triangle would create more obtuse triangles to divide up.
• If one of the angles around the intersection point is $180^\circ$ , we can actually have 4 triangles around an intersection point. This, however, is not viable without creating more centre intersection points as, after the straight line through the intersection cuts off one of the acute corners, the remaining two lines from the intersection must cut the opposite edge to avoid creating obtuse triangles, and then we have 5 obtuse angles due to (1) from each of the constructed lines which must be fixed.

Let me know if there are other ways of dividing the triangle which are not accounted for.