Are the two triangles similar?

Question

A book I was answering was asking for the center of enlargement and the scale factor of the two similar figures.But I don't think the two triangles are similar. They corresponding sides don't have the same ratios. Please correct me if I am wrong.

Answer 1

Answer: Yes

In isosceles $\Delta ABC$ , $BC=2, AB=AC=\sqrt{2^2+1^2}=\sqrt5$

In isosceles $\Delta C'B'A'$ , $A'B'=4, A'C'=B'C'=\sqrt{2^2+4^2}=2\sqrt5$ $$\frac{AB}{C'B'}=\frac{AC}{C'A'}=\frac{\sqrt5}{2\sqrt5}=\frac12, \ \ \frac{BC}{B'A'}=\frac{2}{4}=\frac12$$ $$\frac{AB}{C'B'}=\frac{BC}{B'A'}=\frac{AC}{C'A'}=\frac12$$ $$\therefore \Delta ABC\sim\Delta C'B'A'$$

Answer 2

There are two possibilities. Consider pairwise parallels:

$$ ( AB// A'C'),( AC// B'C'), (BC// B'A') $$

By a simple construction alternate angles an be seen equal. The corresponding included acute angles are equal. So the triangles are similar.

The same is true for the second possibility as well.