There are n line segments drawn from the vertex A to the base BC(not counting the segments AB or AC) as shown. If the total number of triangles of all sizes that can be formed is 66 then the value of n is equal to

There are n+2 lines through A including AB and AC. If BC is one side,then remaining two sides can be selected in $n+2_{C_2}$  ways.there are two more lines parallel to BC.
Total number of triangles =  $n+2_{C_2}$
$n+2_{C_2}$ = 66
 $n+2_{C_2}$= 66 = ${12}_{C_2}$ 
n+2 = 12  $\Rightarrow$ n =10