Let $A^* , B^* , C^* , D^*$ be the points $A , B , C , D$ under the inversion centered at O with lenght R. The image becomes much more simple. Then the claim of this question equals to this claim: "There exist a circle who is tangent to the sides of the quadrilateral $A^*B^*C^*D^*$". Let $O'$ be the point that provides $OA \parallel O'D$ and $OB \parallel O'C$. Because $AB \parallel CD$ the triangles $AOB$ and $DO'C$ are similar. And since $AD \parallel BC$ , then $AD \parallel BC$ $\parallel OO'$. And by using the information $\angle{AOB}+\angle{DOC}=180$ and $\angle{AOB}=\angle{DO'C}$ we get that the quadrilateral $DOCO'$ is cyclic. Then by angle chasing $\angle{ABO}=\angle{DCO'}=\angle{DOO'}=\angle{ADO}$. Since by the properties of inversion the quadrilaterals $A^*ABB^*$ and $AA^*DD^*$ are cyclic thus $\angle{OA^*B^*}=\angle{ABO}=\angle{ADO}=\angle{OA^*D^*}$. But we can do this for $B^*, C^*, D^*$. Thus the angle bisectors of $A^*B^*C^*D^*$ meet at the point O which means there exist a circle that is centered at O and is tangent to the sides of the quadrilateral $A^*B^*C^*D^*$ thus the proof is finished. Note: There must be a more basic one solution but this one is the only one I found. Sorry if there is a mistake I am new around here.