As we know, there are conjectures that are easy to formulate but difficult to prove, and there are conjectures that are easy to prove but difficult to conceive. This conjecture is simple to conceive and probably has a simple proof, however I could not find any reference to it on the Internet so I quote it here:
In the image below, the cyclic crossed quadrilateral A, B, C, D is orthodiagonal, in the sense that the lines passing through the vertices A, C and B, D intersect each other perpendicularly at a point M outside the circle. Prove that the perpendicular to a side from the point M always bisects the opposite side (the intersecting sides AD and BC are considered opposite each other).
Brahmagupta's theorem: https://en.m.wikipedia.org/wiki/Brahmagupta_theorem
In Geogebra, the transition from Brahmagupta's theorem to the above conjecture is done naturally by moving the yellow control point in and out of the circle, respectively: https://www.geogebra.org/m/wthe5qt2
As K bisects AB and $\angle AMB=90^\circ$, $|MK|=|KB|$.
$\angle MBK=\angle MCD$ by inscribed angles on AD.
Therefore $\angle BMK=\angle MBK$.
Let X be a point towards B such that $\angle KMX=90^\circ$, $\angle BMX=\angle MDC$, and therefore MX is parallel to CD.