Exercise on a fixed end Lagrange's MVT Given a function f with derivative on all $[a;b]$ with $f'(a) = f'(b)$, show that there exists $c \in (a;b)$ such that $f'(c) = \frac{f(c)-f(a)}{c-a}$.
This is some kind of MVT with constraint. I have a proof but it uses Darboux's theorem. Can you prove it without using it?

Sketch of the proof I have using Darboux
I suppose first that $f'(a) = 0$ (it's easy to get back to this case with an affine transform) and I set $g(x) = \frac{f(x)-f(a)}{x-a}$. $g'(b) = - \frac{g(b)-g(a)}{b-a} = - g'(d)$ for some $d \in (a;b)$ by Lagrange's MVT. Then $g'(b)$ and $g'(d)$ are either null or have different signs in which case we can find $e \in (d;b)$ such that $g'(e) = 0$. In all cases $g'$ has a zero which is the $c$ we need to find.

(Slight edit: technically we need $h(a) := f'(a)$ in order to make it work. This is because we need continuity for the extreme value theorem, and
$$\lim_{x \to a}\frac{f(x) - f(a)}{x-a} = f'(a). )$$
This problem actually appears to have a fairly elementary answer! (Which is not to say "obvious", clearly.) Define the auxiliary function $$h(x) = \frac{f(x) - f(a)}{x-a}$$ and differentiate by the quotient rule to obtain the following:
$$h'(x) = \frac{f'(x)(x-a) - (f(x)-f(a))}{(x-a)^2}$$
Since $h(x)$ will be continuous on $[a,b]$ and differentiable on $(a,b)$, it will attain an extreme value at a point $c \in (a,b)$ and thus have derivative $0$ there. Thus $f'(c)(c-a) - (f(c) - f(a)) = 0$.
