Given the points $A,B,C,D$ in a straight line $m$ and $A,E,F,G$ in a straight line $n$, how many triangles can be formed with these points? Given the points $A,B,C,D$ in a line $m$ and $A,E,F,G$ in a straight line $n$, how many triangles can be formed with these points?

I've done the following: 


  


I've used the following heuristic: Chose one point on $m$, then another point on $m$ which is different of the previous chosen point. This yields $4*3$, now choose a new point on $n$, this gives me $4*3*3=36$, now choose one point on $n$ and another point on $n$ different of the last point chosen and different of $A$. This gives me $3*2*3=12$ and hence $36+12=48$.

I've also tried to do it in the following way:

$${\text{#points in m} \choose 2}*\text{#points in n (w/o A)}+ {\text{#points in n (w/o A)} \choose 2}*\text{#points in m (w/o A)}$$

Then:

$${4 \choose 2}*3 + {3 \choose 2}*3=27$$

But none of them is right. I may be missing some triangles but I cant see which ones are these.
Outline: First let us count the triangles that have $A$ as a vertex. We need to choose a point other than $A$ on each line.

Next we count the triangles that don't have $A$ as a vertex. We pick $2$ points other than $A$ on one line, and $1$ on the other, or the other way around.
Here is an example, 
If you choose BEGC: one line from BE, one from GC, and one from EG--> you have one more triangle. Do the same  for the rest and I think you may get the result.