It is enough to show that $(AOC)$ pass through midpoint $M$ of $BD$ iff $(A,C;B,D)=-1$. Consider the intersection $P$ of the tangents to $(O)$ at $A$ and $C$. We'll get that $P,A,C,O$ lie on the same circle, so $\angle{PMO}=90^{\circ}$ Since $OM\perp BD$, we get $P,B,D$ collinear and so $(A,C;B,D)=-1$. All the relations we used can be done backward, so we are done.

Let $M,N$ be the midpoints of $AC$ and $BD$. We invert about $\odot O$. This sends $N$ to the intersection of the tangents to the circle at $B,D$. Our condition tells us that $A,C,N'$ are collinear, which implies that the symmedian $AN'$ goes through $C$, or that $ABCD$ is harmonic quad. The symmetry is now pretty nice. We notice that $BD$ is also a symmedian, so it contains the point $M'$ that is the intersection of tangents at $A,C$. This collinearity tells us that $O,B,D,M$ are concyclic by inversion properties.

let $M$ the midpoint of $BD$ .$BM\perp OM$ thus $BD$ cuts $(AOC)$ at the antipode of $O$ i.e. the intersection of its tangents at $A,C$ so it 's the $B$-symmedian hence AC is as well and the circle $BDO $ will pass through its midpoint. RH HAS