Solution:

Let $D^{\prime}$ be the reflection of $D$ with respect to $AB$, and let $E$ be the intersection of line $CD^{\prime}$ with line $AB$. Since the angles $\widehat{BAD}, \widehat{CBA}$ are acute, $E$ lies inside segment $AB$. Moreover, since $C$ and $D^{\prime}$ are equidistant from $AB$, we have $CE = ED^{\prime}$.

Through $E$ draw the lines $r_{1} \parallel AC$ and $r_{2} \parallel AD^{\prime}$, and denote by $G$ the intersection of $r_{1}$ with $AD^{\prime}$, and by $F$ the intersection of $r_{2}$ with $AC$; by construction we get
$$AEF = EAG, \quad CEF = ED^{\prime}G.$$

Similarly, through $E$ draw the lines $r_{3} \parallel BD^{\prime}$ and $r_{4} \parallel BC$, and denote by $M$ the intersection of $r_{3}$ with $BC$, and by $H$ the intersection of $r_{4}$ with $BD^{\prime}$; we get
$$CEM = ED^{\prime}H, \quad BEM = EBH.$$

Finally, let $O, N$ be the reflections of $H, G$ with respect to $AB$, and the choice
$$\begin{gathered} X_{1} = AEF, \quad X_{2} = CEF, \quad X_{3} = CEM, \quad X_{4} = BEM \\ Y_{1} = EAN, \quad Y_{2} = EDN, \quad Y_{3} = EDO, \quad Y_{4} = EBO \end{gathered}$$
verifies the claim.