Tangent at two different points of a simple closed convex curve Let $C$ be a simple closed convex plane curve and let $p$ be line tangent at two points of curve $C$, $t_0 $ and $t_1, t_1\ne t_2$. Show that $C$ contains every point from $[t_1,t_2]$ on $p$, i.e. $p$ is also tangent for every point between $t_1$ and $t_2$ on curve $C$.
My attempt: Let $c$ be the unit length parametrization of our curve with period L and let $t_1,t_2\in [0,L)$ such that $p$ is tangent line at $c(t_1)$ and $c(t_2)$. For sake of contradiction, assume our statement does not hold so there exists a point $t\in (t_1,t_2)$ such that $p$ is not tangent at $c(t)$.
Let $t_0=sup\{t\in (t_1,t_2) \mid \mbox{ p is not the tangent line at c(t)}\}$.
At this point I'm pretty much stuck. I'm trying to get a contradiction with convexity of our curve, so to show that some points of curve lie on different sides of tangent line at $c(t_0)$ .
 A: I'm also trying to work this out, so I'll share what I've got so far. My approach is roughly the same, except I'm using $t_0 = \inf\{t \in (t_1, t_2): p \text{ is not the tangent line at } c(t)\}$. In fact, $p$ is the tangent line at $c(t_0)$. To see this, note that $c'(t) = c'(t_1)$ for all $t \in [t_1, t_0)$, so by continuity of $c'$, $c'(t_0) = c'(t_1)$. Further, $c(t_0) = c(t_1) + \int_{t_1}^{t_0}c'(t) \, dt = c(t_1) + (t_0 - t_1) c'(t_1) \in p$. The goal is to find a $t \in (t_0, t_2)$ such that $c(t_1)$ and $c(t_2)$ are on opposite sides of the tangent line at $c(t)$, i.e. $\langle c(t_1) - c(t), R_{90}(c'(t))\rangle$ and $\langle c(t_2) - c(t), R_{90}(c'(t))\rangle$, where $R_{90}$ is rotation by $90^\circ$, have opposite signs. Such a $t$ will occur "just past" $t_0$.
First, note that $\langle c(t_1) - c(t_0), R_{90}(c'(t))\rangle = \langle c(t_2) - c(t_0), R_{90}(c'(t))\rangle = 0$. Since $t \mapsto \langle c(t_1) - c(t), R_{90}(c'(t))\rangle$ and $t \mapsto \langle c(t_2) - c(t), R_{90}(c'(t))\rangle$ are both smooth, it thus suffices to show that
$$\frac{d}{dt}\langle c(t_1) - c(t), R_{90}(c'(t))\rangle = \langle c(t_1) - c(t), R_{90}(c''(t))\rangle$$
and
$$\frac{d}{dt}\langle c(t_2) - c(t), R_{90}(c'(t))\rangle = \langle c(t_2) - c(t), R_{90}(c''(t))\rangle$$
have opposite signs on $(t_0, t_0 + \varepsilon)$ for some $\varepsilon > 0$.
