How many triangle are there (part 2)?

All possible triangles that can be formed with the blue quadrilateral below as part of its vertex must also contain $1$ of the red segments below, for a total of $5$ possible triangles:

Counting in this fashion, each of the $5$ blue outer polygons have $5$ possible triangles each, each of the $4$ green polygons in the next layer have $4$ possible triangles each, each of the $3$ yellow polygons in the next layer have $3$ possible triangles each, each of the $2$ orange polygons in the next layer have $2$ possible triangles each, and the last red triangle has $1$ possible triangle.

This gives a total of $5^2 + 4^2 + 3^2 + 2^2 + 1^2 = \boxed{55}$ triangles.

First note that any triangle obtained from this figure must

(i) be formed using 3 non-parallel lines, either 2 blue lines and 1 red line, OR 2 red lines and 1 blue line;

(ii) contain either the vertex $A$ or the vertex $B$ .

Let's consider those triangles containing the vertex $A$ . The number of triangles is ${6 \choose 2} +{5 \choose 2}+{4 \choose 2}+{3 \choose 2} +{2 \choose 2}$ . Then consider those triangles containing NO vertex $A$ but vertex $B$ , there are ${5 \choose 2} + {4 \choose 2} +{3 \choose 2} +{2 \choose 2}$ such triangles. Now note that ${{n+1} \choose 2} +{n \choose 2} = n^2$ . Hence there are a total of $5^2+4^2+3^2+2^2+1^2= \boxed{55}$ triangles.

I made a video which is related to this problem, you may check it out.