Avoiding Circular Reasoning: How to Define Congruent Shapes

congruences-geometrygeometryrotations

I apologize for being overly verbose here, but the question I want to know is at the very bottom. I am going to be honest and say I have no idea how to axiomatically handle congruence in geometry and want to understand it.

I have always held SSS, SAS, and CPCTC to be axiomatic where SSS and SAS are definitions to tell whether or not two triangles are congruent. Later, I realized I should not do that here

Why does everyone say two shapes (e.g. triangles here) are congruent iff there is an isometry between the figures? I honestly view the word "isometry" as a made up word describing SSS by a Euclidean metric function that all of the sudden discusses "types of rigid motions". Clearly, if the distances between points are the same in context of triangles, then they satisfy SSS and vice versa.

I am confused why the word isometry is discussed with rotations in particular… What is a rotation and how does that preserve distance here? How can a figure possibly be rotated in space by the means of a function this early on in geometry?

We rotate points in space in terms of sine and cosine. We derive rotational matrices in terms of sine and cosine by double angle formulas. We define sine and cosine in terms of similarity which is done by similarity and axioms using SSS and SAS. We define similarity just like congruence with a scale factor. This leads me to this question again…

Question: How are figures rotated by functions in terms of axioms in the context of isometries?

Best Answer

One way to approach it is to define "reflection" as your fundamental transformation, i.e. for a line $\ell$, there is a transformation $R_\ell$ that preserves collinearity, distance, and angle, but swaps one half-plane for the other.

Then, isometries can be defined as compositions of reflections. Specifically, a rotation about $O$ consists of a consecutively-applied pair of reflections $R_{\ell_1}$ and $R_{\ell_2}$ where $\ell_1$ and $\ell_2$ cross at $O$.

Related Question