Tangents of a curve whose trace remains in a half-plane. I need to prove the following result:

If a curve's trace is contained in a half-plane then the tangents at
  the intersection of the curve with the line defining the half-plane
  $R$ coincide with $R$.

In other words:

If $\alpha:I \to \mathbb{R}^2$ is a regular curve and $H^+,H^-$ denote
  the two closed semiplanes determines by line $R$ then show that the
  points in $t \in \alpha(\mathbb{R}) \cap R$, verify that $R \equiv
 \alpha(t)+L(\alpha'(t))$ the tangent line at $t$.

I have doubts that I have enough tools to prove this result. Could you provide some hints on how to solve it?

Assume that $R$ does not pass through $0$. For any half-plane that does not contain $0$ there exists a unique vector $x \in \mathbb{R}^2$ such that $$y \in H^+ \Leftrightarrow x.y \geq 1$$ and $$y \in H^- \Leftrightarrow x.y \leq 1.$$ If we assume that $\alpha$ lies in $H^+$ and touches the line $R$ at $t_0$ then $x.\alpha(t) \geq 1$ for all $t \in \mathbb{R}$ and $x.\alpha(t_0)=1$. Define the map $f:\mathbb{R} \rightarrow \mathbb{R}, ~ t \mapsto x. \alpha(t)$, then by the linearity of the dot product we can see that $f'(t) = x.\alpha'(t)$. Since $t_0$ is a minimum point of $f$ then $f'(t_0)=x. \alpha'(t_0)=0$. The result now follows as we can define $R$ to be the line $t \mapsto x+ tx^\perp$.
If $R$ contains $0$ then we can define choose $x \in \mathbb{R}^2$ such that $$y \in H^+ \Leftrightarrow x.y \geq 0$$ and $$y \in H^- \Leftrightarrow x.y \leq 0.$$ We now employ the same argument as above.
