Solution We wish to prove $|OE| \cdot |OP| = R^2$. Let $R^2 = |OD| \cdot |OD|$, so we want to show that \begin{align*} |OE| \cdot |OP| &= |OD| \cdot |OD| \\ \iff \frac{|OE|}{|OD|} &= \frac{|OD|}{|OP|}. \end{align*}This would follow immediately if $\triangle OED \sim \triangle ODP$. First we show $O$, $E$ and $P$ are collinear. We have $ABCD$ is an isosceles trapezium, as it's cyclic (well known) then $E, O$ and $P$ lie on the same line by symmetry. Then we have the following. Let $PO$ intersect the circle at $D'$. Then $$ \angle EDO = \angle BDD' = \angle BAD. $$As $DD'$ is a diameter (passes through $O$), we have $\angle D'AD = D'AP = 90^\circ$ by Thales theorem. Then as $AB \perp OP$, let $X$ be the intersection of $PO$ and $AB$ and $Y$ be the intersection of $PO$ and $AD'$. We have $\triangle PAY \sim \triangle AXY$ as shared $\angle AXY$ and right angle, thus $\angle YAX = \angle APX = \angle ODE$. Therefore $\triangle OED \sim \triangle ODP$ due to $\angle OPD = \angle EDO$ and shared angle $\angle EDO$, which implies the result.

Let $AB\cap CD=P_{\infty}$ be the point of infinity along $AB$. It's well known that the inverse of $E$ with respect to $(ABCD)$ is the foot of the perpendicular from $E$ to $PP_{\infty}$. By symmetry, $EP\perp AB$, implying $EP\perp PP_{\infty}$ meaning $P$ must be the inverse of $E$. This immediately gives the conclusion.

$\angle AOC=2\angle B=\pi-\angle APC$, so $P\in (AOC)$. The same $P\in (BOD)$ With inversion about $(O)$ we have $AC\cap BD \mapsto (AOC)\cap (BOD)$, i.e. $E\mapsto P$.

zuss77 wrote: $O$ is intersection of perp bisector of $AD$ and bisector of $\angle APC$, so $P\in (AOC)$. The same $P\in (BOD)$ With inversion about $(O)$ we have $AC\cap BD \mapsto (AOC)\cap (BOD)$, i.e. $E\mapsto P$. I expect that the $OE \cdot OP = R^2$ gives away that there's a clean inversion soln.... If only I knew how to do that

Notice that the circumcircles of triangles $\triangle PAB$ and $\triangle PCD$ are tangent at $P$ since $\overline{AB} \parallel \overline{CD}$, thus $P$ is the Miquel point of $ABCD$, so it's well-known that if $E=\overline{AC} \cap \overline{BD}$, then $OE\cdot OP=R^2$. $\square$

What does OE•OP = R^2 mean in this question?