, 2009

Let $K, L, M, N$ be the points of tangency of the incircle of $ABCD$ to the sides $AB, BC, CD, DA$, respectively, and let $S$ be the foot of the perpendicular from $O$ to the line $PQ$. Since $ABCD$ is a tangential quadrilateral, the point of intersection of the lines $KM$ and $LN$ coincides with the point of intersection of the diagonals $AC$ and $BD$. In other words, $KM \cap LN = \{E\}$.

On the other hand, the points $O$, $K$, $P$, $M$ are concyclic; the points $O$, $L$, $Q$, $N$ are concyclic; and $S$ is the other intersection point of these two circles. In particular, the lines $OS$, $KM$, $LN$ are the radical axes of pairs of these two circles and the incircle of $ABCD$. Therefore, these three lines are concurrent at the radical center of the three circles. In other words, $E$ is the radical center. From this observation it follows that $OE \cdot d = OE \cdot OS = OE \cdot (OE + ES) = OE^2 + OE \cdot ES = OE^2 + LE \cdot EN = OE^2 + r^2 - OE^2 = r^2$.