Question
Show that $\triangle\text{ABC},$ with vertices $A(-2, 0), B(0, 2)$ and $C(2, 0)$ is similar to $\triangle\text{DEF}$ with vertices $D(-4, 0), F(4, 0)$ and $E(0, 4).$
✓
Answer
Using Distance formula
$\text{AB}=\sqrt{(0+2)^2+(2-0)^2}$
$=\sqrt{(2)^2+(2)^2}$
$=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\text{ units}$
$\text{BC}=\sqrt{(2-0)^2+(0-2)^2}$
$=\sqrt{(2)^2+(2)^2}$
$=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\text{ units}$
$\text{CA}=\sqrt{(-2-2)^2+(0-0)^2}$
$=\sqrt{(-4)^2+(0)^2}$
$=\sqrt{16+0}=\sqrt{16}=4\text{ units}$
$\text{DE}=\sqrt{(4+0)^2+(4-0)^2}$
$=\sqrt{(4)^2+(4)^2}$
$=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}\text{ units}$
$\text{EF}=\sqrt{(4-0)^2+(0-4)^2}$
$=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}\text{ units}$
$\text{FD}=\sqrt{(-4-4)^2+(0-0)^2}$
$=\sqrt{64+0}=\sqrt{64}=8\text{ units}$
i.e. $\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{AC}}{\text{DF}}$
$=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{4}{8}=\frac{1}{2}$
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