Answer
See proof
Work Step by Step
Let $g(x)$ and $h(t)$ be the position functions of the two runners and let $f(t) = g(t)-h(t)$.
By hypothesis
$f(0) = g(0)-h(0) = 0$ and
$f(b) = g(b)-h(b) = 0$ where $b$ is the finishing time
Then by the Mean Value Theorem, there is a time $c$ with $0 \lt c \lt b$ such that $f'(c) = \frac{f(b)-f(0)}{b-0} = 0$.
Since $f'(c)$ = $g'(c)-h'(c)$ = $0$
we have $g'(c)$ = $h'(c)$
so at time $c$, both runners have the same speed $g'(c)$ = $h'(c)$.
