Let $\gamma\colon [0,1] \to \mathbb R^2$ (continuous) be a simple closed plane curve and $\,\mathcal C$ its image. Let $x,y$ be some functions such that $\gamma(t)=(x(t),y(t))$. $\gamma$ is said differentiable if $x,y$ are differentiable, $\gamma'(0)=\gamma'(1)$ and $\gamma'(t)$ is never the zero vector.
$\gamma$ is said strictly convex if for any two points on $\,\mathcal C$, the line segment that joins them stays entirely inside the interior of $\,\mathcal C$ (as defined by the Jordan curve theorem), except for the endpoints.
The tangent line to $\gamma$ at $t \in [0,1[$ is the set $$ \{ \gamma(t)+r\gamma'(t) : r \in \mathbb R\} .$$
Is this true?
If $\gamma$ is strictly convex, then any line crosses $\mathcal C$ in at most two points. If $\gamma$ is also differentiable, then a line $s$ crosses $\mathcal C$ in just one point iff $s$ is tangent to $\gamma$. No interior point of $\mathcal C$ are on a tangent line.