Find the number of triangles in the given figure.

Concept: To count triangles in such a figure, first list every triangle that cannot be split further — a “simplest” triangle whose three sides are each a single, unbroken segment of a drawn line. Then check every pair of simplest triangles that share a side: whenever their outer boundary still has exactly three straight sides, that pair combines into one additional, larger triangle. The total number of triangles equals the number of simplest triangles plus the number of these larger, two-piece triangles — no triangle should be counted twice, and no larger combination should be missed.

Applying this to the figure: label the mid-points of the square's four sides as E, F, G and H, and the centre — where the joining segments EG and HF cross — as I, as shown below.

• Four simplest triangles sit at the corners of the square, each cut off by one side of the inner diamond: AEH, EBF, FGC and DGH.

• Four more simplest triangles meet at the centre I, each cut off by the two segments crossing there: EHI, EFI, FGI and GHI.

Each centre triangle shares a side with its neighbour around I, and together their outer boundary runs straight across the centre, forming one larger triangle:

• EHI and EFI together form triangle HEF.

• EFI and FGI together form triangle EFG.

• FGI and GHI together form triangle FGH.

• GHI and EHI together form triangle EGH.

So the figure has 8 simplest triangles and 4 larger triangles built from them, a total of 8 + 4 = 12 triangles.

Cross-check: apart from the square's four sides, the diamond's four sides, and the two segments EG and HF, no other straight line joins two corners of the square (there is no diagonal such as AC or BD in the figure). So no triangle larger than these four combined ones is possible, confirming the count of 12 is complete.