Question

Which theorem or postulate proves that △ABC and △DEF are similar?

Answer 1

Answer: SSS similarity theorem

Step-by-step explanation:

In the given figure we have two triangles ΔABC and ΔDEF.

Since we have given only the side lengths of the triangle.

Thus when we find the ratio of the corresponding sides we get,

$\frac{AB}{DE}=\frac{9}{15}=\frac{3}{5}\\\\\frac{BC}{EF}=\frac{6}{10}=\frac{3}{5}\\\\\frac{AC}{DF}=\frac{12}{20}==\frac{3}{5}\\\\\Rightarrow\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}=\frac{3}{5}$

So by SSS similarity theorem, we have

ΔABC is similar to ΔDEF.

• SSS similarity theorem says that if the lengths of the corresponding sides of two triangles are proportional then the triangles are similar triangles.

Answer 2

3 sides proportional