Minimal perimeter of a triangle

Question

Imagine a triangle with a base $[0, s]$ and a height $h$. ($s, h \gt 0$)

For what orthocentre $x$ does the triangle have a minimal perimeter and how long is it?

Now, the proof starts with:

Assume $(x, h)$ is the third edge point of the triangle (next to $(0, 0$) and ($s, 0$)), then the perimeter is identical with

$$s + \sqrt{x^2 + h^2} + \sqrt{(x - s)^2 + h^2}.$$

How did he conclude that?

Answer 1

Like this?

The two diagonal sides of the triangle are given by Pythagoras' Theorem

Edit:

What if I add the orthocentre in like this? I assume $x$ is the $x$ coordinate of the orthocentre $C$?

Answer 2

$f(z)=\sqrt{h^2+z^2}$ is a convex function on $[0,s]$, hence $$ f(x)+f(s-x) \geq 2\, f\left(\frac{s}{2}\right)$$ and equality is achieved only at $x=\frac{s}{2}$.
With a more geometric approach, let we consider the following configuration: $\ell$ is a line parallel to $AB$ at a distance $h$ from $AB$ and $B'$ is the symmetric of $B$ with respect to $\ell$. For any $C\in\ell$, $$ AC+CB = AC+CB' \geq AB' $$ by the triangle inequality, and equality holds only if the projection of $C$ on $AB$ is the midpoint of $AB$.