Meusnier's theorem states that all curves on a surface $S$ embedded in $\Bbb R^3$ passing through a given point $P$ and having the same tangent $v\in T_PS$ also have the same normal curvature.

I think the same is true for [geodesics torsion.](https://encyclopediaofmath.org/wiki/Geodesic_torsion) See [here](https://en.wikipedia.org/wiki/Darboux_frame#Geodesic_curvature,_normal_curvature,_and_relative_torsion) for my definition of geodesics torsion.

My proof goes as: Take a curve $\gamma : (-\varepsilon, \varepsilon)\to S$ with $\gamma(0)=P,\,\gamma'(0)=v$ and let $u=n\times\gamma',$ where $n$ is the Gauss map. Then 
$$\langle dn(v),n\times v\rangle =\langle (n\circ\gamma)',u \rangle(0)=-\langle n\circ\gamma, u' \rangle (0)=-\langle n\circ\gamma,-\kappa_g v-\tau_g n \rangle(0) =\tau_g.$$
However, the inputs of this formula are independent of the curve. Therefore geodesic curvature depends only on $P$ and $V.$

But this statement (assuming it is correct) and almost one-line proof cannot be found anywhere in the literature. Have I done something wrong?