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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.

enter image description here

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  • $\begingroup$ Yes: Black: base=height=2; Red: base=height=4. $\endgroup$ – David G. Stork May 29 at 2:59
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    $\begingroup$ It's a bit tricky because A, B, C don't correspond to A', B', and C' respectively. But the triangles are similar. ABC ~ C'A'B'. $\endgroup$ – DreiCleaner May 29 at 3:02
  • $\begingroup$ Does that mean BA and B'A' are not corresponding sides? $\endgroup$ – harpey1111 May 29 at 3:04
  • $\begingroup$ It's not true that "$\triangle ABC\sim \triangle A'B'C'$" according to the way the triangles are named. However, it is true that "this triangle is similar to that one". (As David notes, we have isosceles triangles whose base-to-height ratios match.) Without seeing the wording of the problem at hand, it's impossible to say what may be going on here; perhaps the figure is simply mislabeled, or perhaps the author is providing an object lesson about how corresponding elements aren't required to have matchy labels (although matchy labels are certainly helpful). $\endgroup$ – Blue May 29 at 3:05
  • $\begingroup$ Oh. I learned from geometry that similar triangles must be labeled correctly, especially when they are transformed like in transformation lessons $\endgroup$ – harpey1111 May 29 at 3:07
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Answer: Yes

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

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

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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.

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