Here's my solution, finding 35 triangles.

#!/usr/bin/perl

use strict;
use warnings;

#
# Input, lines of the figure.
#
my @l = (
    [qw [A B]],
    [qw [A C F I]],
    [qw [A D G J]],
    [qw [A E]],
    [qw [B C D E]],
    [qw [B F H J]],
    [qw [B I]],
    [qw [E G H I]],
    [qw [E J]],
    [qw [I J]],
);

# Process the lines. Create a datastructure with nodes as keys.
# For each node, record which nodes are reachable in one step,
# and which nodes are passed to get there.
my $g;
foreach my $l (@l) {
    for (my $i1 = 0; $i1 < @$l-1; $i1++) {
        for (my $i2 = $i1+1; $i2 < @$l; $i2++) {
            $$g{$$l[$i1]}{$$l[$i2]} = {map {$_, 1} @$l[$i1+1 .. $i2-1]
+};
            $$g{$$l[$i2]}{$$l[$i1]} = {map {$_, 1} @$l[$i1+1 .. $i2-1]
+};
        }
    }
}

# Find all non-trivial 3-cycles.
local $" = " - ";
my @n = sort keys %$g;
for (my $i1 = 0; $i1 < @n-2; $i1++) {
    for (my $i2 = $i1+1; $i2 < @n-1; $i2++) {
        for (my $i3 = $i2+1; $i3 < @n; $i3++) {
                                         # Form a cycle
            print "@n[$i1,$i2,$i3]\n" if $$g{$n[$i1]}{$n[$i2]} &&
                                         $$g{$n[$i2]}{$n[$i3]} &&
                                         $$g{$n[$i3]}{$n[$i1]}
                                      && # Avoid trivial cycle
                                        !$$g{$n[$i1]}{$n[$i2]}{$n[$i3]
+} &&
                                        !$$g{$n[$i1]}{$n[$i3]}{$n[$i2]
+} &&
                                        !$$g{$n[$i2]}{$n[$i3]}{$n[$i1]
+};
        }
    }
}

__END__
A - B - C
A - B - D
A - B - E
A - B - F
A - B - I
A - B - J
A - C - D
A - C - E
A - D - E
A - E - G
A - E - I
A - E - J
A - F - J
A - G - I
A - I - J
B - C - F
B - C - I
B - D - J
B - E - H
B - E - I
B - E - J
B - F - I
B - H - I
B - I - J
C - E - I
D - E - G
D - E - J
E - G - J
E - H - J
E - I - J
F - H - I
F - I - J
G - H - J
G - I - J
H - I - J
[download]

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