For ΔABC, ∠A = x + 30, ∠B = 2x - 4, and ∠C = 4x. If ΔABC undergoes a dilation by a scale factor of 1 2 to create ΔA'B'C' with ∠A' = 2x + 8, ∠B' = x + 18, and ∠C' = 5x - 22, which confirms that ΔABC∼ΔA'B'C by the AA criterion?

ABC
Solve for x
x + 30 + 2x - 4 + 4x = 180 All triangles have 180°. collect like terms.
7x + 26 = 180 subtract 26 to both sides.
7x = 154 Divide by 7
x = 154/7
x = 22
A = x + 30 = 22 + 30 = 52
B = 2x - 4 = 2(22) + 30 = 44 - 4 = 40
C = 4x = 4*22 = 88

Check
<A + <B + <C = 180
<52 + <40 + <88 = 180°
180 = 180

It checks and we are done with this triangle.

A'B'C' 
2x + 8 + x + 18 + 5x - 22 = 180 Collect the like terms.
8x + 4 = 180 Subtract 4 from both sides.
8x = 180 - 4
8x = 176 Divide by 8
x = 176 / 8
x = 22
 
A' = 2x + 8
A' = 2(22) + 8 = 44 + 8 = 52
B' = x + 18 = 22 + 18 = 40
C' = 5x - 22 = 5*22 - 22 = 110 - 22 = 88

Conclusion
A = A'
B = B'
C = C'

All three angles of one triangle are equal to all three angles of the other. So by AA the two triangles are similar.

Comment
Note: the dilation of 12 has nothing to do with this answer. If you were asked about similarity of sides then the 12 would be something to do with the problem.