Let $A$ be a fixed point in the interior of a circle $\omega$ with center $O$ and radius $r$, where $0<OA<r$. Draw two perpendicular chords $BC,DE$ such that they pass through $A$. For which position of these cords does $BC+DE$ maximize?
$(BC+DE)^2 = BC^2 + DE^2 +2.BC.DE$
$=(AB+AC)^2 + (AD+AE)^2 + 2.BC.DE$
$=AB^2 + AC^2 +AD^2 +AE^2 +2.AB.AC+2.AD.AE+2.BC.DE$
$ \le BD^2+CE^2 + 4\delta_{A/(O)} + \frac{(BC+DE)^2}{2}$
$\Rightarrow (BC+DE)^2 \le 2(BD^2+CE^2)+8\delta_{A/(O)}$
Note that:
$BD^2 + CE^2 = 4r^2(\sin^2 \widehat{BED}+\sin^2 \widehat{EBC})=4r^2(\sin^2\widehat{BED}+\cos^2 \widehat{EBC}) =4r^2$
Therefore,
$BC+DE \le \sqrt{8r^2+8\delta_{A/(O)}} =const$
This sum maximizes
$\Leftrightarrow BC=DE$
$\Leftrightarrow ABCD$ is a isosceles trapezoid and $DE \perp BC$
To construct such this trapezoid, draw a square $OHAK$. The $BC$ would be the chord passes through $A, H$ and $DE$ would be the chord passes through $A, K$