Question

please answer this question ​

Answer 1

Answer:By distance formula ,

In ΔABC  : AB =2√2 , BC= 2√2 , AC=4

In ΔDEF : FE=DE= 4√2 , DF=8

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

Hence the triangles are similar.

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