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My question is raised from this problem:

enter image description here

It's known that the isosceles triangle $AB$ is similar to $CD$. Let's say the ratio of the area of those two triangle is $x\,:\,y$.

My question is, if I draw a cyan line so that I have a triangle $E$ (the big cyan triangle) so that it will similar to the small triangle $D$ (small cyan), will the ratio of $E$ and $D$ be $x\,:\,y$ as well? If yes, could you explain it briefly regarding the proof?

My original problem, I have to find the ratio of the triangle $A$ and $D$ which looks like they're similar as well. However, that's not what I'm going to ask here. The question has already written above. Mentioning my original problem, probably you could understand why I ask this.

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From the problem statement we can say wlog the base of $AB$ is $\sqrt{x}$ and the base of $CD$ is $\sqrt{y}$.

Given that $E$ and $D$ are similar, by the SSS criterion, their ratio is also $\sqrt{x}:\sqrt{y}$ as in the figure below, where $\sqrt{x}$ is the side of E and $\sqrt{y}$ is the side of D

enter image description here

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    $\begingroup$ Thank you for your comment. Could you give me a link about what LLL criterion is? I have tried to search on Google and couldn't find one. $\endgroup$
    – user516076
    Dec 29, 2021 at 1:52
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    $\begingroup$ LLL, commonly known as SSS $\endgroup$ Dec 29, 2021 at 1:55
  • $\begingroup$ LOL. I haven't used this criterion since high school and was thinking "it was length-length-length?" $\endgroup$
    – Rol
    Dec 29, 2021 at 1:59
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    $\begingroup$ You state that there is a ratio $x:y$ between the lengths, but you haven't gone into whether it's the same ratio as exists between the larger triangles. Also, the asker used $x:y$ for the ratio of areas, not of lengths. $\endgroup$ Dec 29, 2021 at 2:00
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    $\begingroup$ @TimPederick thank you! I fixed these issues $\endgroup$
    – Rol
    Dec 29, 2021 at 2:13
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Yes, similarity preserves the same proportions between parts as it does the whole (angles equal, lengths scaled $1:n$, areas scaled $1:n^2$).

I'm not familiar with any succinct statement of this fact, unlike CPCTC for congruence, but the principle is the same. $AB$ is the image of $CD$ under a certain combination of translation, rotation, and scaling. The same translation, rotation and scaling will produce an image of $D$ that is similar: $E$.

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