Bulgarian National Olympiad

Denote by $M$ and $N$ the tangent points of the incircle of $ABCD$ with $AB$ and $CD$, respectively. It follows from $\angle BAD+\angle ADC>180^{\circ }$ that $AB\parallel CD$ and $\angle MIN<180^{\circ }$. Also, the equalities $IM=IN$, $\angle IMX=\angle INY$ and $IX=IY$ show that $△IMX\cong △INY$, implying $\angle IYN=\angle IXM$. If $X\in BM$ and $Y\in DN$ (or $X\in AM$ and $Y\in CN$) then the equality $\angle IYN=\angle IXM$ implies $AB\parallel CD$, a contradiction. Therefore $X\in XB$ and $Y\in NC$. It follows from $AXYD$ that $$\angle AXI=\angle DYI=180^{\circ }-\frac{\angle A}{2}-\frac{\angle D}{2},$$ giving $\angle AIX=\frac{\angle D}{2}$ and $\angle DIY=\frac{\angle A}{2}$. Therefore $△AIX\sim △IDY$, which implies that $AX\cdot DY=IY\cdot IX$. Analogously, $△BIX\sim △ICY$, i.e. $BX\cdot CY=IY\cdot IX$. Hence $AX\cdot DY=BX\cdot CY$.