How to show that there exists $\xi \in (a,b)$ such that $f'(\xi)=\frac{f(\xi)-f(a)}{\xi-a}$

Question

Given that $f(x)$ is differentiable on the closed interval $[a,b]$ and $f'(a)=f'(b)$.How can I prove that there exists $\xi \in (a,b)$ such that$$f'(\xi)=\frac{f(\xi)-f(a)}{\xi-a}$$ I have tried to let $$ F(x)=\begin{cases} \frac{f(x)-f(a)}{x-a}&a<x\leq b \\ f'(a)&x=a \\ \end{cases} $$ And it's easy know that when $a<x < b$,$$F'(x)=\frac{f'(x)(x-a)-[f(x)-f(a)]}{(x-a)^2}$$ If $F(a)=F(b)$,we can easily prove the statement with Rolle's theorem.But I don't think it's possible to prove $F(a)=F(b)$ here. So I am totally at a loss what to do now.