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.
A: The geometric solution by @Jack D'Aurizio is beautiful !  An analytic solution may still feel safer:
An affine transformation may map the given lines onto x- and y- axes, and point $\ P\ $ to $\ (1\,\ 1).\ $ Since the affine maps preserve the proportions of areas, and the proportion of the lengths of any two intervals lying on the same straight line, the questions is reduced to this special case. (To compute things in general is possible but less elegant).
Let $\ A=(0\,\ a)\ $ and $\ B=(b\,\ a)\ $ be as above, i.e. $\ (1\,\ 1)\ $ is between $A$ and $B.\ $ Then $\ a>1\ $ and $\ b>1\ $, and
$$ a\cdot b\ =\ a+b $$
Thus, triangle $AOB$ has minimal area (which happens to be equal $2)\ $ for $\ a=b=2,\ $ which is the required answer; indeed, the double triangle's area is at least $4$:
$$ a\cdot b\ - 4\ =\ a+b-4\ =\ (a-2) + (\frac a{a-1} - 2)\ =
                   \ \frac{(a-2)^2}{a-1}\ \ge\ 0 $$
Thank you.
A: Thought I'd try some arguments from pre calculus.
Geometrically we know area cannot follow below $p_xp_y$ were $P=(p_x,p_y)$.
Suppose you have a line through $P$ that intersects the $x$ and $y$ axes. 
As you rotate the line through P, you can change the angles at which the line intersects the x and y axes. Rotate clockwise, the angle with respect to the x axis increases and the angle with respect to the y axis decreases. The opposite happens on clockwise rotations. 
The area of a sector of a circle is $A_{sec}=\frac{1}{2}r^2d\theta$.
Let $r1$ = distance from P to the vertex on x axis and $r2$=distance from P to the vertex on the y axis.
The change in area of a sector is approximately the change in area of the triangle as the  line rotates. The change in areas near the vertices have opposite signs. So the total change in area is approximately $$\delta A\approx\frac{1}{2}(r_2^2-r_1^2)\delta\theta$$
Far away from P, the change in area by rotation is roughly $\frac{1}{2}x^2 \frac{p_x}{x}\delta\theta$, or vice versa substituting x for y. So the further away, the faster the change in area, overall proportional to the intercept. So there is no maximum area. 
We have a lower bound an no upper bound. As with the vertex of a parabola representing a projectile under gravity, we expect there to be a place where the change in area with respect to angle changes from positive to negative or vice versa hitting zero in between. 
From the above sector formula, we know this can only happen if $r_1=r_2$, which means P is the midpoint of the hypotenuse. 
My Thales' Theorem, the  midpoint of the hypotenuse lies at the center of a circle passing through every vertex of the right triangle. 
So the x and y intercepts of the line are solutions to the equation $(x-p_x)^2+(y-p_y)^2=p_x^2+p_y^2$
