A diagonal is a line that connects any 2 vertices of the polygon, right? 

So, combinatorially, now many pairs of vertices can you choose in a 12 sided polygon?

Note that for a diagonal, the order of the vertices does not matter. 

i.e. if a diagonal connects vertices A and B, it is the same as the 
diagonal connecting B to A and you would count it only once. 

So the answer is "12 choose 2" or C(12,2) which is 12*11/1*2 = 66. 

But here is the catch - this set of combinations would also include the 
12 edges of the polygon 

i.e. if it had vertices from A to L, the edges AB, BC, CD etc. also would be counted in the above formula. 

However, they are not diagonals but edges.

Hence we have to subtract 12 from the above answer to get the number of  diagonals.

In general, for a polygon of n vertices (n >= 3), the number of diagonals would be given by the formula

C(n,2) - n =  upon simplification.

Hope this helps. Do mail me if it's not clear.

:)