Conic from Parallel Chords
What Might This Be About?
Two circles intersect at points $A$ and $B.$ Chords $EF$ on one and $GH$ on the other are parallel.
Prove six points $A,B,E,F,G,H$ lie on a conic.
The easiest way to prove the statement is to show that it is equivalent to its inverse. This is done in a reasonably standard manner by reductio ad absurdum.
Given six points on a conic, $A,B,E,F,G,H.$ Such that $ABFE$ and $EBHG$ are cyclic quadrilaterals. Then $EF\parallel GH.$
The problem has been posted by Dao Thanh Oai at the CutTheKnotMath facebook page.