Mongolian Mathematical Olympiad

$$\begin{gathered} (AB+x{)}^2+(CD+x{)}^2-(AB+x)(CD+x)=A{D}^2⟺ \\ \Rightarrow AB\cdot x+CD\cdot x-AB\cdot CD=0. \end{gathered}$$ On the other hand, $(AB+CD-x{)}^2=A{B}^2+C{D}^2+x^2-2(AB\cdot x+CD\cdot x-AB\cdot CD)=A{B}^2+C{D}^2+x^2=A{D}^2$ and we get $AB+CD=AD+x=AD+BC$. It means in the quadrilateral $ABCD$ can be inscribed a circle.