Consider the figure consisting of a square, its diagonals, and the segments joining the midpoints of opposite sides. The total number of triangles of any size in the figure is 
$\mathrm{(A) \ 10 } \qquad \mathrm{(B) \ 12 } \qquad \mathrm{(C) \ 14 } \qquad \mathrm{(D) \ 16 } \qquad \mathrm{(E) \ 18 }$
[asy]draw((0,0)--(2,2)); draw((0,2)--(2,0)); draw((0,0)--(0,2)); draw((0,2)--(2,2)); draw((2,2)--(2,0)); draw((0,0)--(2,0)); draw((1,0)--(1,2)); draw((0,1)--(2,1));[/asy]
There are 8 little triangles, 4 triangles with twice the area, and 4 triangles with four times the area of the smaller triangles. $8+4+4=16\Rightarrow \mathrm{(D)}$