Suppose you have A,B,C,D four points in harmonic range and O, P are the midpoints of AB and CD respectively. Prove that $OP^2=PC^2+OB^2$.
I would guess that this is a sort of corollary to the theorem that says:
If $(ACBD)$ is a harmonic range and if $O$ is the midpoint of $AB$ then $OB^2=OC\cdot OD.$
But I'm not sure
Let $F$ on the circle with diameter $AB$ be such that $FC$ is the angle bisector of $\angle AFB$. Since $(ACBD)$ is a harmonic range, $FD$ is the exterior angle bisector at $F$ of $\triangle AFB$. Consequently $\angle AFD=135^\circ$ and $FC\perp FD$.
It follows that $\angle OAF+\angle PDF=45^\circ.$ Hence $$\angle FOP+\angle FPO=90^\circ,$$ and we have $$OP^2=OF^2+PF^2=OB^2+PC^2.$$