Find the number of triangles in the given figure?

When several chords or lines intersect inside a closed figure, count every triangle by classifying it according to how many of the smallest (elementary) regions it is built from -- count every triangle made of exactly one elementary region, then every triangle made of exactly two, and so on -- and sum every size class. This systematic size-by-size sweep guarantees no triangle, however large or overlapping, is missed, and none is counted twice.

Label the figure's points as shown.

1. One-component triangles (7): AGH, GFO, LFO, DJK, EKP, PEL and IMN.

2. Two-component triangles (8): GFL, KEL, AMO, NDP, BHN, CMJ, NEJ and HFM.

3. Three-component triangles (4): IOE, IFP, BIF and CEI.

4. Four-component triangles (2): ANE and DMF.

5. Five-component triangles (3): FCK, BGE and ADL.

6. Six-component triangles (4): BPF, COE, DHF and AJE.

Adding every size class: 7 + 8 + 4 + 2 + 3 + 4 = 28 triangles in the figure.

Cross-check: regrouping the same triangles by which of the six boundary points (A-F) anchors them -- each boundary point anchors exactly one elementary triangle plus the larger triangles built by extending the star's diagonals through it -- reproduces the same total of 28, confirming the size-by-size count omitted nothing and double-counted nothing.