How many non-congruent triangles have vertices at three of the eight points in the array shown below?
This problem can be solved by manual counting together with the principle of symmetry.
First we only need to consider the base of the triangle is on the bottom four dots (use the principle of symmetry to eliminate the congruent case when the base is on the top four dots). Then we can do casework by the length of the base:
Hence the answer is $3+3+2=\boxed{8}$.