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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.

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    try to use google, the MVT is proven very often – 2017-01-14
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    Have you read the exercise? It's not the MVT at all, I used this term because the result look a bit like the MVt. – 2017-01-14
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    The statement given as a question is named Flett MVT and it was proved in 1958. So, the proof can be found in a literature. In particular, Riedel and Sahoo wrote a book devoted to Mean Value Theorems. This is "Mean value theorems and functional equations". The proof of Flett's Theorem could be found there. – 2017-01-14

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The function $$ h(t) = \left\{ \begin{matrix} \frac{f(t)-f(a)}{t-a} & t\in (a,b] \\ f'(a) & t=a \end{matrix} \right. $$ is continuous on $[a,b]$ whence attains its max and min values in the closed interval $[a,b]$. It is differentiable on $(a,b]$ with a derivative $$ h'(t) = \frac{f'(t)(t-a)-(f(t)-f(a)) }{(t-a)^2}$$ If suffices therefore to see that $h$ attains an extremal value at a point $c$ in the open interval $(a,b)$.

Suppose first that

$$h(b)=\frac{f(b)-f(a)}{b-a} < h(a)=f'(a)=f'(b)$$

Let $c\in[a,b]$ be such that $h$ attains its minimum value at $c$. As $h(b)a$. But since $h'(b)> 0$ we also have that $c$ must be strictly smaller than $b$. So $c\in (a,b)$ and we have $h'(c)=0$ by minimality.

A similar argument works for the reversed inequality (take $c$ to be a point where the max of $h$ is attained).

Finally, if there is equality then either $f$ is linear and every $c\in (a,b)$ works, or you may take $c$ to be a point where either a max or a min different from $h(b)=h(a)$ is attained (this is in fact just the usual MVT).

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This paper https://arxiv.org/pdf/1309.5715.pdf contains two nice proofs of Flett's MVT, which is a statement of your question. Of course, Riedel and Sahoo book I pointed in the comment is also good.

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(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 $f'(c)(c-a) - (f(c) - f(a)) = 0$.