Question

Show that $△ABC$ with vertices A(-2, 0), B(0, 2), and C(2, 0) is similar to $△DEF$ with vertices D(-4, 0), F(4, 0) and E(0, 4)

Answer 1

To prove that $\Delta ABC\sim \Delta DEF$

we need to show that $\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$

We know that the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is,

$\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

In $\Delta$ABC vertices are A(-2, 0), B(0, 2), and C(2, 0).

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

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

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

In $\Delta$DEF vertices are D(-4, 0), F(4, 0) and E(0, 4).

$DE=\sqrt{(0--4{)}^2+(4-0{)}^2}=\sqrt{(4{)}^2+(4{)}^2}=\sqrt{32}=4\sqrt{2}$

$EF=\sqrt{(4-0{)}^2+(0-4{)}^2}=\sqrt{(4{)}^2+(4{)}^2}=\sqrt{32}=4\sqrt{2}$

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

$\frac{AB}{DE}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}\frac{BC}{EF}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}\frac{AC}{DF}=\frac{4}{8}=\frac{1}{2}$

$i.e,\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$

$\therefore \Delta ABC\sim \Delta DEF$