Existence and uniqueness of a point with horizontal tangent in a convex curve I had a look on the proof by E. Schmidt of the Schur's Theorem about arcs of convex curves. It states the following:

Let $C$ and $C'$ be two arcs of the same length with the endpoints $a,b,a',b'$ respectively, $\;d=\overline{ab}$ and $\;d'=\overline{a'b'}$ denoting the distance of the endpoints; furthermore let $k(s)$ and $k'(s)$ denote the respective curvatures of $C$ and $C'$, where the parameter $s$ is the arclength on $C$ and $C'$ measured from $a$ and $a'$ respectively.
  If $C$ is a plane curve and together with the chord connecting its endpoints forms a simple closed convex curve, and if at every point $s$, $0\leq s\leq l$, $k'(s)\leq k(s)$ holds, then $d'\geq d$.

The first claim in the proof is that since $C$ possesses a continuously turning tangent there exists a point $s_1$, $0<s_1<l$, where the direction of the tangent
to $C$ is parallel to the chord through $a$ and $b$.
Also, by hypothesis $C$ together with the chord connecting $a$ and $b$ form a simple closed convex curve; therefore the angle enclosed by the tangent at $C(s)$ with the line through $a$ and $b$ is a monotonic function of $s$ and its variation on the arcs $\;0\leq s\leq s_1$, $\;s_1\leq s\leq l\;$ is no greater than $\;\pi$.

Well, I certainly urderstand these two facts and I can picture them in my mind. Despite of this, I'm not able to give a rigorous proof.
My thought is that one can consider the tangent vector continuously at every point $s$, $\;0\leq s\leq l$ and use an argument similar to Rolle's Theorem (considering as the referencial axis the chord through $a$ and $b$) to see that there exist this point $s_1$ where the tangent is parallel to the chord. I'm not sure if this is correct and still it remains to prove the uniqueness.
To see that the variation of the angle in the arc is no greater than $\pi$: convexity implies that the total curvature is $2\pi$, so "without" the chord the arc $C$ has a variation in the angle lower or equal to $\pi$ (shouldn't be actually equal to $\pi$?).
Any thoughts or ideas about these arguments are very welcome.
 A: I think I can prove this facts with rigorous arguments, as I wanted.
First, we consider $g$ the curve formed by the arc $C$ and line through $a$ and $b$. Orient this curve so that the segment $ab$ lie on the $x$ axis.
Rolle's Theorem can be used if we write $g(s)=(x(s),y(s))$ and observe that the function $y:[0,l]\to\mathbb{R}$ is continuous and differentiable in $[0,l]$ and it holds that $y(0)=y(l)=0$. Then, there exist some $s_0\in (0,l)$ such that $y'(s_0)=0$.
Now, $g'(s)=T'(s)=(x'(s),y'(s))$ is the tangent vector of $g$ at $s$. Then, $\;T'(s_0)=(x'(s_0),0)$ is a vector with the $y$ component equal to zero, so parallel to the $x$-axis (and also to the line through $a$ and $b$).
To prove uniqueness of $s_0$ I need to add as hypothesis that the curvature $k(s)$ is always positive (in most of the newer versions of this theorem that I have seen in these days, it is done this way). Now suppose there are more than just one point with this property. We could face two possibilities:


*

*There are two (or more) critical points of $y$. Then automatically the convexity condition would fail.

*There is a 'plain' in the curve, where the tangent is parallel in more than one point. In this case, the condition of $k(s)>0$ fails and we are done.

