State whether the following statement is true or false. Justify your answer.
$\triangle ABC$ with vertices $A(-2, 0), B(2, 0)$ and $C(0, 2)$ is similar to $\triangle DEF$ with vertices $D(-4, 0), E(4, 0)$ and $F(0, 4)$.

(TRUE) True.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
For $\triangle ABC$:
$AB = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2 + 0^2} = 4$
$BC = \sqrt{(0 - 2)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$
$AC = \sqrt{(0 - (-2))^2 + (2 - 0)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$
For $\triangle DEF$:
$DE = \sqrt{(4 - (-4))^2 + (0 - 0)^2} = \sqrt{8^2 + 0^2} = 8$
$EF = \sqrt{(0 - 4)^2 + (4 - 0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
$DF = \sqrt{(0 - (-4))^2 + (4 - 0)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$
Comparing the ratios of corresponding sides:
$\frac{AB}{DE} = \frac{4}{8} = \frac{1}{2}$
$\frac{BC}{EF} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2}$
$\frac{AC}{DF} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2}$
Since $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = \frac{1}{2}$,by the $SSS$ similarity criterion,$\triangle ABC \sim \triangle DEF$.