For ΔABC, ∠A = 4x - 10, ∠B = 5x + 10, and ∠C = 7x + 20. If ΔABC undergoes a dilation by a scale factor of 1 3 to create ΔA'B'C' with ∠A' = 2x + 10, ∠B' = 8x - 20, and ∠C' = 10x - 10, which confirms that ΔABC∼ΔA'B'C by the AA criterion?

We know that the angles of a triangle sum to 180°.  For ΔABC, this means we have:
(4x-10)+(5x+10)+(7x+20)=180

Combining like terms,
16x+20=180

Subtracting 20 from both sides:
16x=160

Dividing both sides by 16:
x=10
This means ∠A=4*10-10=40-10=30°; ∠B=5*10+10=50+10=60°; and ∠C=7*10+20=70+20=90.

For ΔA'B'C', we have
(2x+10)+(8x-20)+(10x-10)=180

Combining like terms, 
20x-20=180

Adding 20 to both sides:
20x=200

Dividing both sides by 20:
x=10

This gives us ∠A'=2*10+10=20+10=30°; ∠B'=8*10-20=80-20=60°; and ∠C'=10*10-10=100-10=90°.

Since the angle are all congruent, ΔABC~ΔA'B'C' by AAA.