<s>[INST] Why does rotation preserve angles and distance in the Euclidean plane? An angle in radians is a distance moved around a unit circle. A rotation of a point around a center of rotation, moves the point a distance around a circle around the center that goes through the point. The distance is given by the angle of the rotation multiplied by the distance from center.
So why do rotations preserve distance and angles when the distance measure is Pythagorean.
 [/INST] A linear algebra approach:
Any linear transformation of $\mathbb{R}^n$, such as a rotation about the origin, can be represented with an $n \times n$ matrix $A$, where $A\mathbf{v}$ is the output when the transformation acts on a vector $\mathbf{v}$.  Any counterclockwise rotation of $\mathbb{R}^2$ about the origin looks like $\left[ \begin{array}{ccc}
\cos(\theta) & -\sin(\theta) \\
\sin(\theta) & \cos(\theta) \end{array} \right]$ where $\theta$ is the angle of rotation.  See my post here for a justification of this fact.
First, let us investigate $\|\mathbf{v}\|$ and $\|A \mathbf{v}\|$.  If $\mathbf{v} = \left[ \begin{array}{ccc}
a \\
b \end{array} \right]$, then $\|\mathbf{v}\| = \sqrt{a^2 + b^2}$.  You can check that $A \mathbf{v} =  \left[ \begin{array}{ccc}
\cos(\theta)a - \sin(\theta)b \\
\sin(\theta)a + \cos(\theta)b \end{array} \right]$.  We have: 
$$\|A \mathbf{v}\| = \sqrt{\Big( \cos(\theta)a - \sin(\theta)b \Big)^2 + \Big( \sin(\theta)a + \cos(\theta)b \Big)^2}$$
$$ \quad \quad  \quad \ = \sqrt{\Big( \sin^2(\theta) + \cos^2(\theta) \Big)a^2 + \Big(\sin^2(\theta) + \cos^2(\theta) \Big)b^2}$$
And this reduces to $\sqrt{a^2 + b^2} = \|\mathbf{v}\|$ due to the Pythagorean identity: $\sin^2(\theta) + \cos^2(\theta) = 1$.  Thus, the length of any vector is preserved under any rotation.

To see why rotations preserve angles, use the fact that the cosine of the angle between unit vectors $\mathbf{v}$ and $\mathbf{w}$ is equal to their dot product $\mathbf{v} \cdot \mathbf{w}$.  So let $\mathbf{v} = \left[ \begin{array}{ccc}
a \\
b \end{array} \right]$ and $\mathbf{w} = \left[ \begin{array}{ccc}
c \\
d \end{array} \right]$ be generic unit vectors; you will find that $\mathbf{v} \cdot \mathbf{w}$ and $A\mathbf{v} \cdot A \mathbf{w}$ are equal (following reduction of the latter using trig identities).
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