# If ratio of sides of two triangles is constant then the triangles have the same angles

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?

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 $$\color{blue}{\alpha=\alpha'\; \text{ similarly }\; \beta=\beta'\; \text{ and }\; \gamma=\gamma'}$$

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