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Let $\alpha : (a,b) \rightarrow \mathbb{R}^2$ be a smooth map(infinitely differentiable).

Show that if the chord length $\Vert{\alpha(s)-\alpha(t)}\Vert$ depends only on $|s-t|$, then it is a line or a part of a circle.

It comes from Shifrin's differential geometry notes.

Here is my attempt:

Since the chord length only depends on $|s-t|$, for $s-t=\delta, -\delta$, the chord length should be same.

$\Vert \alpha(s+\delta) -\alpha(s) \Vert = \Vert \alpha(s) - \alpha(s-\delta) \Vert$

Squaring both sides and viewing both sides as an inner product of themselves, respectively.

And expanding Taylor series and discarding terms with higher degree than 2 of $\delta$'s.

Then I arrived at $\alpha(s)' \cdot \alpha(s)'' =0$.

And I have no idea to do. Could you give me a hint?

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