Number of Triangles in this figure How many triangles are in this figure? I suspect there are $45$, but I would like someone to confirm or correct me. I arrived at $45$ after manually counting them. If anyone has a formula by which this number can be obtained, please share.

 A: I get
$$\binom83-\binom53-\binom33-\binom33-1=56-10-1-1-1-\boxed{43}.$$
A triangle is formed by $3$ lines intersecting in $3$ different points. Your figure has $8$ lines, so $\binom83=56$ potential triangles. Since each pair of lines intersects in the figure, we only have to subtract the number of concurrent triples of lines: $\binom53=10$ at B, $\binom33=1$ at A, $\binom33=1$ at C, and $1$ at G.
A: I get only $43$ triangles. Here's how I did the counting: consider the collection of sets of points on each segment
$$S=\{ AKDHC, CEB, BFA, ALGJE, CIGMF, BMLK, BGD, BJIH \}$$
(here for simplicity I'm denoting the points of a segment as a concatenated string).
Now make a graph $g$ where vertices are the points $A,\dots, M$ and put an edge if the points are on a same segment (loop over $S$ and for each pair on a segment put an edge).
Number of closed $3$-walks on this graph gives the number of "triangles" but some of these are degenerated, i.e. all the points lie an a line. But these are easily counted as
$$\sum_{L\in S} { |L| \choose 3} = 41$$
So the answer is $1\over 6$(trace of third power of the adjacency matrix of $g$) minus $41$
$$
\frac{1}{6} tr\left(
\displaystyle \left(\begin{array}{rrrrrrrrrrrrr}
0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 0 \\
1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 1 \\
1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\
1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 \\
1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 1 \\
1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & 1 \\
1 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & 1 & 0 & 0 & 1 & 0 \\
1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 \\
1 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 1 \\
0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 1 & 0
\end{array}\right)^3
\right)
- \sum_{L\in S} { |L| \choose 3} = 43
$$
Here are the triangles that I get by iterating all closed $3$-walks, accepting only non-degenerated and considering permutations the same:
ABC
ABD
ABE
ABG
ABH
ABJ
ABK
ABL
ACE
ACF
ACG
ADG
AFG
AHJ
AKL
BCD
BCF
BCG
BCH
BCI
BCK
BCM
BDH
BDK
BEG
BEJ
BEL
BFG
BFI
BFM
BGI
BGJ
BGL
BGM
BHK
BIM
BJL
CDG
CEG
CHI
CKM
GIJ
GLM

So I get

*

*$15$ triangles that include the point $A$

*$22$ that include $B$ but not $A$

*$4$ that include $C$ but not $A$ or $B$

*$2$ that don't include any of $A,B,C$
Can you spot the $2$ that I've missed or you have over-counted?
