Triangle product puzzle

Label the missing triangles $D$ , $E$ , and $F$ , as shown below:

Then $2ED = 12$ and $2EF = 16$ , which simplifies to $ED = 6$ and $EF = 8$ . Since every number is a positive integer greater than $1$ , $E$ must be a factor of $6$ and $8$ and greater than $1$ , which means $E = 2$ . Since $ED = 6$ and $EF = 8$ and $E = 2$ , $D = 3$ and $F = 4$ .

We also know that $2AF = B$ , and since $F = 4$ , $B = 8A$ . We are given that $A + B = 45$ , and substituting $B = 8A$ gives $9A = 45$ , which means $A = 5$ . Since $A = 5$ and $B = 8A$ , $B = 40$ .

Since $C = 12 \cdot 16 \cdot B$ and $B = 40$ , $C = 7680$ .

Therefore, $25CA^{-1}B^{-2} = 25 \cdot 7680 \cdot 5^{-1} \cdot 40^{-2} = \boxed{24}$ .

We have $8B= A\implies A+B=45 \implies 9B=45 \implies B=5$

Then $A =45-5=40 \implies C=12\times 16\times 5= 960$

The answer is $\frac{960}{40}=24.$

I think the letters A and B must be switched in the equations.