Suppose you have a closed differentiable plane curve. Are the tangent lines to the curve at the most distant points on the curve always parallel? What if we assume that the curve is convex?
I don't see why this should be the case, but it also seems like it might be the kind of weird geometric property that unexpectedly turns out to be true.
Let $\gamma : [0,1] \to \mathbb{R}^2$ be our closed differentiable planar curve with $\gamma(0) = \gamma(1) = x_0$. We wish to find the $s<t$ such that $$F(s,t) := \|\gamma(s) - \gamma(t)\|^2$$ is maximized. Take the derivative with respect to $s$ and $t$: we find that $$\nabla F(s,t) = 2\bigg( (\gamma(s) - \gamma(t)) \cdot \gamma'(s), (\gamma(s) - \gamma(t))\cdot -\gamma'(t)\bigg).$$ At an extremal point, the gradient must be 0: this happens when $s = t$, or at points $s < t$ where $\gamma(s), \gamma(t)$ are farthest away from each other (locally.)
The gradient implies that both $\gamma'(s), \gamma'(t)$ are orthogonal to the vector $\gamma(s) - \gamma(t)$: in $\mathbb{R}^2$, this is exactly what you wanted to prove, because the orthogonal complement of a vector in $\mathbb{R}^2$ is one dimensional. In $\mathbb{R}^3$ or greater, however, we don't have such a guarantee.