Calculus: The Language Of Change (2005) by David W. Cohen, James M. Henle. pp. 827-829. The original colored in just blue. I annotated and added more colors.
Pls see below. Why do $\color{pink}{h}, \color{forestgreen}{v}, \alpha$ never enter or show up in the Mean Value Theorem Proof below? It feels wrong that $\color{pink}{h} = \color{forestgreen}{v}\cos\alpha$ (Exercise 4 on p. 829) transforms $\color{pink}{h}$ in Fig. 2 to $\color{forestgreen}{v}$ in Fig. 3, but the Mean Value Theorem Proof never features $\color{pink}{h}, \color{forestgreen}{v}, \alpha$.
As I understood, comparing the picture from the exercise with the picture from the proof of the theorem at p. 828, $h$, $v$, and $\alpha$ from the exercise, in the proof of the theorem are called the perpendicular distance (which is the distance between a current point $f(t)$ on a graph and the chord), the vertical distance (between the current point and its horizontal projection into the chord), and the angle of the elevation of the chord, respectively. In the proof we want to maximize the perpendicular distance $h$. But $h$ is equal to the product of the vertical distance $v$ by a $\cos \alpha$. Since the chord endpoints $(c,f(c))$ and $(d,f(d))$ are fixed, $\cos \alpha=\tfrac{d-c}{\sqrt{(d-c)^2+(f(d)-f(c))^2}}$ is a constant. So we are maximizing the vertical distance $v$. Moreover, later the vertical distance is denoted by $s(t)=f(t)-q(t)$.