The curve in Figure 1.7.3(a) represents the heights $h\left(s\right)\,s\in \left[0\,1\right]$.

The black line represents a span $s\=1\/2$. The endpoints sit at two equal heights.

The endpoints of the black line are solutions of the equation $h\left(a\+1\/2\right)\=h\left(a\right)$ , $0\le a\le 1\/2$, or the alternative equation

Define $f\left(a\right)\=h\left(a\+1\/2\right)-h\left(a\right)$

If $f\left(0\right)\=0$, then $h\left(1\/2\right)\=h\left(0\right)$, so the heights are equal for $s\=0$ and $s\=1\/2$.

If $f\left(0\right)\ne 0$, then $f\left(1\/2\right)\=h\left(1\right)-h\left(1\/2\right)\=h\left(0\right)-h\left(1\/2\right)\=-\left(h\left(1\/2\right)-h\left(0\right)\right)\=-f\left(0\right)$
Assuming $h\left(s\right)$ is continuous, so $f\left(a\right)$ is continuous, and $f\left(0\right)\cdot f\left(1\&sol;2\right)<0$, the Intermediate Value theorem can be applied to $f$ and there is an $\stackrel{\&Hat;}{a}\ne 0$ for which $f\left(\stackrel{\&Hat;}{a}\right)\&equals;0\&equals;h\left(\stackrel{\&Hat;}{a}\&plus;1\&sol;2\right)-h\left(\stackrel{\&Hat;}{a}\right)$. Thus, the heights are the same at $\stackrel{\&Hat;}{a}$ and $\stackrel{\&Hat;}{a}\&plus;1\&sol;2$.