By the mean value theorem, there exists $c \in (a,b)$ s.t. : $$0=\frac {f(b)-f(a)}{b-a}=f'(c).$$
Knowing that $f'$ is continuous, by IVT, there exists $x_1 \in (a,c)$ such that $f'(x_1)=\dfrac{f'(a)}2$. By the same reasonning, there exists $x_2 \in (c,b)$ such that $f'(x_2)=\dfrac{f'(b)}2$.
Since $f'(a)=f'(b)$, $f'(x_1)=f'(x_2)$ and $x_1 \
eq x_2$ since they belong to disjoints intervals.
Note that we could have taken any value between $f'(a)$ and $0$. Choosing $\dfrac{f'(a)}2$ and $\dfrac{f'(b)}2$ was just for convenience.