$f: [a,b] \to \mathbb{R}$, $f(a) = f(b)$. Show $\exists c \in [a, \frac{a+b}{2} ]$ with $f(c)=f(c+\frac{b-a}{2})$

Question

Let $f$ be a real valued and continuous function over $[a,b]$ with $a <b$ such that $f(a)=f(b)$ , My question here is :

Question (claim): How to show that : there exists $c \in [a, \frac{a+b}{2} ]$ such that : $f(c)=f(c+\frac{b-a}{2})$.

Note: I have tried intermediate values but I didn't succeed.

Answer 1

Consider a function $g: [a, \frac{a+b}{2}] \to \mathbb{R}, g(x) = f(x) - f\left(x+\frac{b-a}{2}\right)$. You want to show that this function has a zero somewhere in its domain. What is $g(a)$? What is $g\left(\frac{a+b}{2}\right)$?