I was working through another problem which reduced to this one, and as the solution and its geometric interpretation was so simple, it got me wondering whether this is provable using geometry.
Please do let me know if this is a well-known problem or a duplicate. I tried to find this problem somewhere but searching for geometric setups is difficult if you don't already know its name.
The problem:
Let $\ell$ be a line and $O$ a point on it. Furthermore, let $A$ be point outside of the line such that $OA\not\perp\ell.$ Find the point $P$ on the line $\ell$ which minimises the ratio of lengths $AP/PO$.
A sketch of my analytic solution (if it helps):
WLOG let $O=(0,0)$, $A=(a,0)$ and $\ell\colon y=kx, k\neq 0.$ Now, a general point on the line $\ell$ is $(x, kx)$, so the problem reduces to finding the $x$ which minimises the function $f\colon \mathbb R\to \mathbb R,$ $$f(x)=\frac{\sqrt{(x-a)^2+k^2x^2}}{\sqrt{(k^2+1)x^2}}.$$ By standard methods of differential calculus we get this $x$ to be $x=a.$ I.e. the point $P$ is such that $PA\perp AO$.
So is there a geometric way (without calculus) to prove (and hopefully at the same time construct the point $P$) that the point $P$ minimises the ratio?
Let $P \in \ell$ be such that $AP\perp OP$.
Case 1 ($OP'> OP$)
Produce $AP$ to $P''\in (OAP')$. By similarity $APP' \sim OP'P'$, and by the fact that $OP''$ is a diameter of $(OAP)$ we have $$\frac{\overline{AP}}{\overline{OP}} = \frac{\overline{AP'}}{\overline{OP''}}<\frac{\overline{AP'}}{\overline{OP'}}.$$
Case 2 ($OP'< OP$)
I leave it to you as an exercise. Use the Figure below, where $AP'$ is produced to $P''\in (OAP)$.