Can you do a mathematical proof that triangles with the same angles have proportional sides

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MegaMoh

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In school, we took similarity, which is basically if you have 2 triangles and both of them have the same angles, they are similar, and they both must be either a scaled up/down version of the other. Similarity is then used throughout math a lot because of the side proportionality. It makes sense looking at it that triangles with the same angles are scaled versions of each other, but not because it "looks right" it is; I thought that there must be a mathematical proof for it(that doesn't use concepts derived from it!).

So does any of you know a mathematical proof that proves that a triangle that has all three angles "congruent" to the angles of another triangle, that means that there is a fixed number you multiply by a side of one of the triangles to get a side of the other? I think using trigonometric function, in this case, would be a cheat because they depend on similarity as well, but I'm not sure. Any help is appreciated.
 
I would probably use a coordinate geometry approach. Suppose we have 3 points in the plane of the form:

[MATH]\left(x_i,y_i\right)[/MATH] where \(i\in\{1,2,3\}\)

Now suppose we generate another 3 points as follows:

[MATH]\left(kx_i,ky_i\right)[/MATH] where \(k\ne0\in\mathbb{R}\)

Using the definition of slope, can you show the new triangle implied by this second set of points must have the same interior angles, and using the distance formula can you show that the sides of this triangle have all been scaled in length relative to the original triangle by the factor \(k\)?
 
I would think that the law of sines or the law of cosines would be enough. Both those theorems are proved using the theorem of Pythagoras.
 
Well, as far as I'm concerned, why d'heck prove something EVIDENT??!!
 
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