Show that ΔABC with vertices A(–2, 0), B(0, 2) and C(2, 0) is similar to ΔDEF with vertices D(–4, 0), F(4, 0) and E(0, 4).

In ΔABC, the coordinates of the vertices are A(–2, 0), B(0,2), C(2,0). 

$\mathrm{AB}=\sqrt{(0+2)^{2}+(2-0)^{2}}=\sqrt{8}=2 \sqrt{2}$

$\mathrm{CB}=\sqrt{(0-2)^{2}+(2-0)^{2}}=\sqrt{8}=2 \sqrt{2}$

$\mathrm{AC}=\sqrt{(2+2)^{2}+(0-0)^{2}}=4$

In ΔDEF, the coordinates of the vertices are D(–4, 0), E(4, 0), F(0, 4).

$\mathrm{DF}=\sqrt{(4+4)^{2}+(0-0)^{2}}=8$

$\mathrm{FE}=\sqrt{(0-4)^{2}+(4-0)^{2}}=4 \sqrt{2}$

$\mathrm{DE}=\sqrt{(0+4)^{2}+(4-0)^{2}}=4 \sqrt{2}$

Now, for ΔABC and ΔDEF to be similar, the corresponding sides should be proportional. 

So, $\frac{\mathrm{AC}}{\mathrm{DF}}=\frac{\mathrm{BC}}{\mathrm{FE}}=\frac{\mathrm{AB}}{\mathrm{DE}}$

$\Rightarrow \frac{4}{8}=\frac{2 \sqrt{2}}{4 \sqrt{2}}=\frac{2 \sqrt{2}}{4 \sqrt{2}}$

$\Rightarrow \frac{1}{2}=\frac{1}{2}=\frac{1}{2}$

Since, the corresponding sides are proportional

Therefore, given two triangles are similar.