Points $X$ and $Y$ inside the rhombus $ABCD$ are such that $Y$ is inside the convex quadrilateral $BXDC$ and $2\angle XBY = 2\angle XDY = \angle ABC$. Prove that the lines $AX$ and $CY$ are parallel.
S. Berlov
$\angle XBY=\frac{_1}{^2}\angle ABC=\angle DBC$ $\Longrightarrow$ $\angle ABX=\angle DBY.$ Similarly, we have $\angle ADX=\angle BDY.$ If $Z$ denotes the isogonal conjugate of $X$ WRT $\triangle ABD,$ we have $\angle DBZ=\angle ABX=\angle DBY$ and $\angle BDZ=\angle ADX=\angle BDY$ $\Longrightarrow$ quadrilateral $BYDZ$ is a kite $\Longrightarrow$ $Z$ is reflection of $Y$ across $BD.$ Thus, by obvious symmetry $ACYZ$ is an isosceles trapezoid with legs $AZ,CY$ $\Longrightarrow$ $\angle ACY=\angle CAZ=\angle CAX$ $\Longrightarrow$ $AX \parallel CY.$
We have $\angle{ABX}\equiv \angle{YBD}$ and analogues. By trigonometric Ceva in
$\triangle{ABX}$ and $\triangle{CBD}\Rightarrow
\dfrac{\sin\angle{BAX}}{\sin{\angle{DAX}}}$$=\dfrac{\sin{\angle{DCY}}}{\sin{\angle{BCY}}}$, so $\angle{BAX}\equiv\angle{DCY}\Rightarrow
AX||CY$ $\square$