2023 WSMO Accuracy Round Problems/Problem 6

Problem

In quadrilateral $ABCD,$ there exists a point $O$ such that $AO = BO = CO = DO$ and $\angle(AOB)+\angle(COD) = 120^{\circ}.$ Let $K,L,M,N$ be the foot of the perpendiculars from $A$ to $BD,$ $B$ to $AC,$ $C$ to $BD,$ and $D$ to $AC.$ If $[ABCD] = 20,$ find $\left([KLMN]\right)^2.$

Solution