$\bigtriangleup$ABC with vertices A(–2, 0), B(2, 0) and C(0, 2) is similar to $\bigtriangleup$DEF with vertices D(–4, 0), E(4, 0) and F(0, 4).

Solution.

$Distance\, Formula= \sqrt{\left ( {x_{2}-x_{1}} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
In  $\bigtriangleup$ABC,
$AB= \sqrt{\left ( 2+2 \right )^{2}+\left ( 0-0 \right )^{2}}= \sqrt{\left ( 4 \right )^{2}}= 4$
$BC= \sqrt{\left ( 0-2 \right )^{2}+\left ( 2-0 \right )^{2}}= \sqrt{\left ( 4+4 \right )}= \sqrt{8}= 2\sqrt{2}$
$AC= \sqrt{\left ( 0+2 \right )^{2}+\left ( 2-0 \right )^{2}}= \sqrt{\left ( 4+4 \right )}= 2\sqrt{2}$
In $\bigtriangleup$DEF,
$DE= \sqrt{\left ( 4+4 \right )^{2}+\left ( 0-0 \right )^{2}}= \sqrt{\left ( 8 \right )^{2}}= 8$
$EF= \sqrt{\left ( 0-4\right )^{2}+\left ( 4-0 \right )^{2}}= \sqrt{16+16}= \sqrt{32}= 4\sqrt{2}$
$DF= \sqrt{\left ( 0+4\right )^{2}+\left ( 4-0 \right )^{2}}= \sqrt{16+16}= 4\sqrt{2}$
$Here,\frac{AB}{DE}= \frac{BC}{EF}= \frac{AC}{DF}$
$\frac{4}{8}= \frac{2\sqrt{2}}{4\sqrt{2}}= \frac{1}{2}$,
Hence, $\bigtriangleup$
ABC is similar to $\bigtriangleup$DEF i.e $\bigtriangleup ABC\sim \bigtriangleup DEF$