Solution: Define a function \( g: [a, \frac{a+b}{2}] \to \mathbb{R} \) by:
\[
g(x) = f(x) - f\left(x + \frac{b-a}{2}\right).
\]
Since \( f \) is continuous on \([a, b]\), \( g \) is continuous on \([a, \frac{a+b}{2}]\). Now observe that:
\begin{align*}
g(a) & = f(a) - f\left(a + \frac{b-a}{2}\right) \\
& = f(a) - f\left(\frac{a+b}{2}\right) \\
& = f(b) - f\left(\frac{a+b}{2}\right).
\end{align*}
Similarly,
\begin{align*}
g\left(\frac{a+b}{2}\right) & = f\left(\frac{a+b}{2}\right) - f\left(\frac{a+b}{2} + \frac{b-a}{2}\right) \\
& = f\left(\frac{a+b}{2}\right) - f(b) .
\end{align*}
Thus,
\[
g(a) = -g\left(\frac{a+b}{2}\right).
\]
If \( g(a) = 0 \), then \( c = a \) satisfies the condition. Otherwise, \( g(a) \) and \( g\left(\frac{a+b}{2}\right) \) have opposite signs. By the
Intermediate Value Theorem, there exists \( c \in \left(a, \frac{a+b}{2}\right) \) such that \( g(c) = 0 \). This implies:
\[
f(c) = f\left(c + \frac{b-a}{2}\right),
\]
as required.