Hello, Krizalid!
I have a rather klunky proof . . .
Prove that two lines are perpendicular if the product of their slopes is $\displaystyle 1$
Without trig, vectors . . . just using basic analytic geometry tools.
The first problem is creating a right angle.
I used that fact that an angle inscribed in a semicircle is a right angle. Code:

* * * P
*  o(x,y)
*  o o*
* o o*
o  o
* o  o*
B o+o A
(r,0)*  *(r,0)

*  *
*  *
*  *
* * *

$\displaystyle P(x,y)$ is a point on the semicircle: .$\displaystyle y \:=\:\sqrt{r^2x^2}$ .[1]
The slope of $\displaystyle PA$ is: .$\displaystyle m_{PA} \:=\:\frac{y}{xr}$ .[2]
The slope of$\displaystyle PB$ is: .$\displaystyle m_{PB} \:=\:\frac{y}{x+r}$ .[3]
Substitute [1] into [2]: .$\displaystyle m_{PA} \:=\:\frac{\sqrt{r^2x^2}}{xr} \:=\:\frac{\sqrt{(rx)(r+x)}}{(rx)} \:=\:\sqrt{\frac{r+x}{rx}}$
Substitute [1] into [3]: .$\displaystyle m_{PB} \:=\:\frac{\sqrt{r^2x^2}}{x+r} \:=\:\frac{\sqrt{(rx)(r+x)}}{r+x} \:=\:\sqrt{\frac{rx}{r+x}} $
Therefore: .$\displaystyle \left(m_{PA}\right)\left(m_{PB}\right) \;=\;\sqrt{\frac{r+x}{rx}}\cdot\sqrt{\frac{rx}{r+x}} \;=\;1$