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If $\triangle ABC$ and $\triangle A'B'C'$ are a pair of triangles such that

$$\dfrac{|AB|}{|A'B'|}=\dfrac{|BC|}{|B'C'|}=\dfrac{|AC|}{|A'C'|}$$

then

$$\triangle ABC \sim \triangle A'B'C'$$

I have shown that the converse is true, could I use that to prove the statement in the title?

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If $\triangle ABC$ and $\triangle A'B'C'$ are a pair of triangles such that $$\dfrac{|AB|}{|A'B'|}=\dfrac{|BC|}{|B'C'|}=\dfrac{|AC|}{|A'C'|}$$

is equivalent to

$\triangle ABC$ has the sides $a, b, c$ and $\triangle A'B'C'$ has the sides $a·k, b·k, c·k$ for some $k\in \mathbb R\;\;$

In virtue of the similarity criterion sss, we obtain, as desired, that $$\triangle ABC\sim \triangle A'B'C'$$


Now, why does this work? Observe, that in virtue of the Law of Cosines

$$\cos{\alpha}=\frac{b^2+c^2-a^2}{2bc}$$ and $$\cos{\alpha'}=\frac{b^2k^2+c^2k^2-a^2k^2}{2bk\cdot ck}=\frac{b^2+c^2-a^2}{2bc}$$ And since $f(x)=\cos(x)$ is injective in $(0, \pi)$, we know that $$\color{blue}{\alpha=\alpha'\; \text{ similarly }\; \beta=\beta'\; \text{ and }\; \gamma=\gamma'}$$

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  • $\begingroup$ Is there a way to do it without the use of trigonometry? $\endgroup$ Mar 30, 2019 at 17:44
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    $\begingroup$ There is in fact; see here $\endgroup$
    – Dr. Mathva
    Mar 30, 2019 at 17:47
  • $\begingroup$ I used the Law of Cosines because it's much faster ;) $\endgroup$
    – Dr. Mathva
    Mar 30, 2019 at 17:48

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