If we have two surfaces in $\mathbb{R}^3 ,$ say $S_1, S_2 $. And we have an isometry $f:S_1 \rightarrow S_2$. Then we have two curves $\alpha $ and $\beta $ on $S_1$ that intersect at $t=0$.
I am trying to show that the angle between the curves $\alpha $ and $\beta $ on $S_1$ is the same as the angle between $f(\alpha) $ and $f(\beta) $ on $S_2$.
The angle between them, $\theta $, on $S_1$ is given by $$\cos \theta =\frac{\alpha '\cdot \beta'}{||\alpha' ||||\beta '||}$$
The angle between the transformed curves $\phi $ is given by $$\cos \phi =\frac{f(\alpha ')\cdot f(\beta)'}{||f(\alpha)' ||||f(\beta) '||}$$
but I don't know how to show they're the same. I can't see where I can use the fact that $f$ is an isometry.
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