suppose i have a smooth path $\gamma:[a,b] \rightarrow \mathbb{R}^2$, also $\gamma '$ exists and $\gamma'(t)\neq0 \space \forall t\in (a,b)$. suppose that $\gamma(a)\neq \gamma(b)$ we want to show that there exists $c\in[a,b]$ such that $\gamma'(c) = \lambda(\gamma(b)-\gamma(a))$ for some $\lambda$.
at first I thought about using lagrange multipliers because of that $\lambda$, but couldnt use it because the range of $\gamma$ is $\mathbb{R}^2$
then i thought maybe looking at the angle of $\gamma'(t)$ and then use the lagrange theorem for but i dont know that $\gamma(a)\neq 0$
okay so suppose that $A=(0,0)$ and $B=(k,0)$ then since $\gamma (t)= \begin{bmatrix} \gamma_1(t) \cr \gamma_2(t)\end{bmatrix}$ we can look only at $\gamma_2(t)$ and since $\gamma_2(a) = \gamma_2(b)= 0$ and $\gamma_2$ is continous on $[a,b]$ and differntiable on $(a,b)$ we can use roll's theorm therefore exists $c\in(a,b)$ such that $\gamma'_2(c)= 0 = \lambda(\gamma_2(b)-\gamma_2(a))$ it even works for any $\lambda$.
so what happends if the path doesnt start at $(0,0)$ and ends on the X axis? we just shift the entire path to $(0,0)$ by subtracting $\gamma(a)$ and rotating by multiplying by the rotating matrix $R(-\varphi)$ while $\phi$ is the arguemt then it we look $f(t)=R(-\varphi)\cdot [\gamma(t)-\gamma(a)]$ and we can look again at the second coordinate of $f$. this time there will be a little problem when $\varphi = 90,0$ but whit a little bit of diving to cases it is solvable.
so far I have looked at my full solution and it seems to work just that its not an elegant solution.