I have a question about the relationship between two different formulations of the notion of boundary convexity, in the sense of Riemannian geomety. Let $M$ be an $n$-dimensional manifold with boundary embedded in $\mathbb{R}^{n+1}$, and assume that $M$ is equipped with the metric induced by its embedding. From what I've seen, it is standard to say that $\partial M$ is convex iff, given $x\in\partial M$, there is a neighborhood $N$ of $x$ such that any two points of $N\cap\partial M$ are joined by a geodesic of $M$. Several sources I've read indicate that this is equivalent to requiring that $\partial M$ has non-negative curvature everywhere with respect to the inward normal.
My main question is if this definition of convexity is equivalent to the following condition: given $x\in\partial M$, there is an affine hyperplane $H$ in $\mathbb{R}^{n+1}$ tangent to $\partial M$ but not to $M$ at $x$ and a neighborhood $N$ of $x$ such that $H\cap N\cap\mbox{Int}(M)=\emptyset$. This seems to me to be true on an intuitive level, but I'm not sure how to go about proving it. I'm primarily interested in the cases $n=2$ and $n=3$.
Update: I think I've found a partial answer in the cases $n=2$ and $n=3$. The above condition implies that $\partial M$ has non-negative curvature with respect to the inward normal, where curvature is taken to be mean curvature in the case $n=3$ (when $\partial M$ is a knotted surface). This is nothing more than a direct consequence of the definitions of curvature in these cases- I was overthinking this. In the case $n=3$, the latter condition does not need to be satisfied at points of $\partial M$ where the mean curvature is zero, so the conditions aren't equivalent in general.