The product of the lengths of two line segments that belong to Newton line I am looking for the proof of the following claim:

Consider a family of bicentric quadrilaterals with the same inradius length and the same distance between incenter and circumcenter. Denote by $P$ and $Q$ the midpoints of the diagonals, and by $I$ the incenter. Then, $|PI| \cdot |QI|$ has the same value for all quadrilaterals in the family.


The GeoGebra applet that demonstrates this claim can be found here.
 A: Let the incircle be the unit circle. Let $A'$, $B'$, $C'$, $D'$ be the points where this is tangent to $AB$, $BC$, $CD$, $DA$, and let them have coordinates $(\cos \alpha, \sin \alpha)$, $(\cos \beta,\sin \beta)$, $(\cos \gamma, \sin \gamma)$ and $(\cos \delta, \sin \delta)$.
Then the quadrilateral $IA'BB'$ has two right angles, so $\angle A'BB' = \pi - \angle B'IA' = \pi - \beta + \alpha$. Likewise, the quadrilateral $IC'DD'$ has $\angle C'DD' = \pi - \delta + \beta$. But these two angles are opposite angles in a cyclic quadrilateral, so they sum to $\pi$. Thus $(\pi - \beta + \alpha) + (\pi - \delta + \beta) = \pi$, and $\delta = \pi + \alpha - \beta + \gamma$.
Now
$$A = \left(
\frac{\sin\alpha-\sin\delta}{\sin(\alpha-\delta)},
-\frac{\cos\alpha-\cos\delta}{\sin(\alpha-\delta)}
\right)$$
$$B = \left(
\frac{\sin\beta-\sin\alpha}{\sin(\beta-\alpha)},
-\frac{\cos\beta-\cos\alpha}{\sin(\beta-\alpha)}
\right)$$
$$C = \left(
\frac{\sin\gamma-\sin\beta}{\sin(\gamma-\beta)},
-\frac{\cos\gamma-\cos\beta}{\sin(\gamma-\beta)}
\right)$$
$$D = \left(
\frac{\sin\delta-\sin\gamma}{\sin(\delta-\gamma)},
-\frac{\cos\delta-\cos\gamma}{\sin(\delta-\gamma)}
\right)$$
Letting $\theta=\alpha-\beta$ and $\phi=\beta-\gamma$, we have:
$$|PI|^2=\frac14(A+C)\cdot(A+C)=\frac{1-\sin\theta\, \sin\phi}{(\sin\phi)^2}$$
$$|QI|^2=\frac14(B+D)\cdot(B+D)=\frac{1-\sin\theta\, \sin\phi}{(\sin\theta)^2}$$
And after finding the circumcenter $O$,
$$|IO|^2=\frac{1-\sin\theta\, \sin\phi}{(\sin\theta)^2(\sin\phi)^2}$$
This leads to $$|IO|^2=|PI||QI|+|PI|^2|QI|^2$$
or in unit-free terms
$$|IO|^2=|PI||QI|+|PI|^2|QI|^2/\text{inradius}^2$$
which establishes the claim.
