If every normal section of a surface is a geodesic, then for every point of the surface, the curvature is the same in any direction. [closed]

How should I show that if every normal section of a surface in $$\mathbb{R}^3$$ is a geodesic, then for every point of the surface, the curvature is the same in any direction?

I would like some hints. Thank you!

Sketch: Assume $$p=(0,0,0)$$ is a point on the surface $$S$$ and the unit normal vector to $$S$$ at $$p$$ is $$(0,0,1)$$. Take a ball of radius $$r>0$$ small enough on the tangent plane at $$S$$ (the $$X,Y$$-plane). Note that if you go $$r$$ distance from $$p$$, along $$S$$, in any direction, to say $$x$$, the normal vector at $$x$$ is in the plane passing through $$Z$$-axis and $$x$$ (uniqueness of geodesics given a direction and at a point).
Now consider the curve on $$S$$ which is the set of all points at Riemannian distance $$r$$ from $$p$$. Consider the function which assigns to each point on this curve its euclidean distance from the $$X,Y$$-plane. Note that the vectors tangent to this curve are perpendicular to planes containing the base point and the $$Z$$-axis by the observation above. This means that the function considered above is a constant. This is true for any $$r$$. Check that this gives local symmetry (rotations in the $$X,Y$$-plane send geodesics from $$p$$ to geodesics at $$p$$). The claim follows from this.
• When considering the curve which is the set of all points at distance $r$ from $p$ (second paragraph), do you mean Euclidean distance? or intrinsic distance? Nov 30, 2020 at 6:33