To find the length of AC, we must form a proportional relation between the triangles. Observe that EF corresponds to BC, and DE corresponds to AB.
![(EF)/(BC)=(DE)/(AB)](https://img.qammunity.org/2023/formulas/mathematics/college/xio4yhxfbrjmrjifycuihxdcf1gnspa1ur.png)
Replacing all the given expressions, we have.
![(2x+10)/(12)=(56)/(16)](https://img.qammunity.org/2023/formulas/mathematics/college/kg3c2x1gc8ok4jz5nwjozzg5v7izyzf5tu.png)
Now, we solve for x.
![\begin{gathered} 2x+10=56\cdot(12)/(16) \\ 2x+10=42 \\ 2x=42-10 \\ 2x=32 \\ x=(32)/(2) \\ x=16 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4xn7kzmd1zqm06r3wfsrx9681w2l805s0v.png)
Then, we use this value to find AC.
![AC=2x+8=2(16)+8=32+8=40](https://img.qammunity.org/2023/formulas/mathematics/college/e8gitesn0mzkagannyqvvpmwdto0h9azx4.png)
Therefore, AC must be 40 units long to prove that the triangles are similar.