Let $M \subset \mathbb{R}^{3}$ be a regular surface such that the intersection of $M$ with the plane $$ \{(x,y,z) \in \mathbb{R}^{3} : z=0\} $$ is the image of a curve parametrized by arc length $\alpha$.
I am trying to prove that if $F:M\rightarrow \mathbb{R}^{3}$ defined by $F(x,y,z) = (x,y,-z)$ is an isometry of $M$ then the curve $\alpha$ must be a geodesic of $M$.
I would like to prove that the vector $(0,0,1)$ belongs to the tangent plane $T_{q}M$ with $q\in M \cap P$. I have tried many things but I'm not making any progress. So I would really appreciate some help.