Question

Triangle ABC with vertices A (-2,0),B(2,0),C(0,2) is similar to triangle DEF D(-4,0),E(4,0),F(0,4) . Prove it.

Answer 1

Answer:

Hello Bhakti Gulati here hope you found this answer easy:)

Step-by-step explanation:

Answer 2

Answer:

All the corresponding sides are proportional, therefore triangle ABC and DEF are similar triangles.

Step-by-step explanation:

The vertices of triangle ABC are A (-2,0),B(2,0),C(0,2) and the vertices of triangle DEF are D(-4,0),E(4,0),F(0,4).

Distance formula:

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

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

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

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

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

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

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

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

Since all the corresponding sides are proportional, therefore triangle ABC and DEF are similar triangles.

Hence Proved.