Problem 3

Triangles $AWZ$ and $CYX$ have the same apex angle $\angle A=\angle C$ and the same inradius $r_1=r_3$, so they share the same decreasing relation between their two legs; likewise $\triangle BXW$ and $\triangle DZY$ share $\angle B=\angle D$ and $r_2=r_4$. If $AZ\ne CX$, assume without loss of generality $AZ>CX$. Then monotonicity at $A,C$ gives $CY>AW$; subtracting from $AB=CD$ gives $BW=AB-AW>CD-CY=DY$; monotonicity at $B,D$ then gives $DZ>BX$; subtracting from $DA=BC$ gives $CX=BC-BX>DA-DZ=AZ$, contradicting $AZ>CX$. Hence $AZ=CX$ and $AW=CY$ (and consequently $DZ=BX$ and $BW=DY$). Thus $\triangle AWZ\cong\triangle CYX$ and $\triangle BXW\cong\triangle DZY$ by SAS, giving $WZ=YX$ and $WX=ZY$, so $WXYZ$ is a parallelogram.