Let $\alpha,\beta,\gamma,\delta$ be the angles at vertices $A,B,C,D$, respectively. Without loss of generality we can assume that the angles $\alpha,\beta$ are acute. In this case both circumcenter $O$ and incenter $I$ lie inside $\triangle ABK$, where $K$ is the intersection point of the diagonals.
We have:
$$\angle BID=2\pi-\frac12\beta-\gamma-\frac12\delta=\frac\pi2+\alpha,
\quad \angle BOD=2\alpha.$$
Let $\rho(P,P_1P_2)$ denote the distance of a point $P$ to a line drawn through the points $P_1$ and $P_2$. Then
$$
\frac{\rho(O,BD)}{\rho(I,BD)}=\frac{A_{BOD}}{A_{BID}}
=\frac{OB\cdot OD\cdot\sin(\angle BOD)}{IB\cdot ID\cdot\sin(\angle BID)}
=\frac{R^2\sin2\alpha}{\frac{r}{\sin\frac\beta2}\frac{r}{\sin\frac\delta2}\sin\left(\frac\pi2+\alpha\right)}=\frac{R^2}{r^2}\sin\alpha\sin\beta.\tag1
$$
The same result will be obtained for the ratio of the distances to the diagonal $AC$. Thus:
$$
\frac{\rho(O,BD)}{\rho(I,BD)}=\frac{\rho(O,AC)}{\rho(I,AC)}\iff
\frac{\rho(O,AC)}{\rho(O,BD)}=\frac{\rho(I,AC)}{\rho(I,BD)},
$$
which means that the points $K,I,O$ are collinear. As a side result we obtain:
$$\begin{align}
\frac{KO}{KI}&=\frac{R^2}{r^2}\sin\alpha\sin\beta\\
&=\frac{(ab+cd)(ac+bd)(ad+bc)}{16\,abcd}\frac{(a+c)(b+d)}{abcd}\frac{2\sqrt{abcd}}{ad+bc}\frac{2\sqrt{abcd}}{ab+cd}\\
&=\frac{(ac+bd)(a+c)(b+d)}{4abcd},\tag2
\end{align}
$$
where formulas from wikipedia page were used to express the radii and angles via the sides of the quadrilateral.
