Derivative parallel to secant line on a curve Let $\alpha: I \subset \mathbb{R} \rightarrow \mathbb{R}^2$ a regular curve and $a < b \in I$, such that $\alpha(a) \neq \alpha(b)$. Show that there exists $s_0 \in (a,b)$, such that $\alpha'(s_0)$ is parallel to the line segment joining the extreme points $\alpha(a)$ and $\alpha(b).$
I know that this a very intuitive geometric fact, my idea was to use the mean value theorem to find the $s_0$, but I don't know how to apply it because the curve does not assume values in $\mathbb{R}$. This together with my lack of ability to write equations of geometric entities made me be stuck on this very simple problem. A detailed solution will be appreciated, I need gain ability in geometry, but first I need a sample on how to think geometrically.
 A: Let $\alpha(t)=(\alpha_{x}(t),\alpha_{y}(t))$
The vector from $\alpha(a)$ to $\alpha(b)$ is
$$(\alpha_{x}(a)-\alpha_{x}(b),\alpha_{y}(a)-\alpha_{y}(b))$$
A vector perpendicular to it is
$$(\alpha_{y}(a)-\alpha_{y}(b),-\alpha_{x}(a)+\alpha_{x}(b),)$$
Consider
\begin{align}
\phi(t)
&\triangleq \alpha(t) \cdot (\alpha_{y}(a)-\alpha_{y}(b),-\alpha_{x}(a)+\alpha_{x}(b),) \\
&= \alpha_{x}(t)[\alpha_{y}(a)-\alpha_{y}(b)]+\alpha_{y}(t)[-\alpha_{x}(a)+\alpha_{x}(b)]\\ \end{align}
Computation (!) shows that $\phi(a)-\phi(b) = 0$
Hence by Rolle's theorem there exist $s \in (a,b)$ such that
\begin{align}
 \phi'(s) = 0
&\iff \alpha'_{x}(s)[\alpha_{y}(a)-\alpha_{y}(b)]+\alpha'_{y}(s)[-\alpha_{x}(a)+\alpha_{x}(b)] = 0 \\ 
&\iff (\alpha'_{x}(s),\alpha'_{y}(s)) \cdot (\alpha_{y}(a)-\alpha_{y}(b),-\alpha_{x}(a)+\alpha_{x}(b)) = 0 \\ 
&\iff \alpha'(s) \cdot (\alpha_{y}(a)-\alpha_{y}(b),-\alpha_{x}(a)+\alpha_{x}(b)) = 0 \\ 
\end{align}
This means that $\alpha'(s)$ is parallel to the vector (hence the line segment) from $\alpha(a)$ to $\alpha(b)$.
