Q:

Right triangle minimum area problem without calculus

Consider two perpendicular lines $\ell_1$ and $\ell_2$ which intersect at $O$. There is another fixed point $P$ somewhere. We want to choose points $A$ and $B$ on the lines $\ell_1$ and $\ell_2$ such segment $AB$ contains $P$ and such that we minimize the area of triangle $\triangle OAB$. I think that we should choose the points where the circle centered at $P$ with radius $OP$ intersects the lines. How should you prove this?
Here is a geogebra to play with (you can move point $A$ around until it intersects the circle): https://www.geogebra.org/m/wxQXjdmt

There have been several questions along the same lines but all have been with an explicit point $P$ (given coordinates), and they use calculus: Minimum or maximum area of the triangle formed by a linear function and the axes and Optimization problem: given that a line passes through $(4,3)$ and it forms a triangle with x and y axis, find minimum area and Find equation of line such that area formed by line & positive coordinate axis is minimal for example.
I would prefer an example that does not do a "set coordinates and bash with calculus" approach because that is ugly. This seems like a nice geometric fact, so a geometric proof with limited calculus would be best.

A:

Fold it! By denoting through $P_1$ and $P_2$ the projections of $P$ on the lines $\ell_1,\ell_2$, you can easily notice that the area of $OAB$ is at least twice the area of the rectangle $OP_1 PP_2$, unless $P$ is the midpoint of $AB$:

It follows that the minimum area is attained when $A$ and $B$ lie on the circle centered at $P$ with radius $PO$, as conjectured.

