4. Given that $A_{1}, B_{1}, C_{1}, D_{1}$ are points on the four sides of square $A B C D$. The following are two propositions about it:
(1) If $A_{1} B_{1} C_{1} D_{1}$ is a rectangle with unequal adjacent sides, then the rectangle $A_{1} B_{1} C_{1} D_{1}$ must have a side parallel to one of the diagonals of the square $A B C D$;
(2) If $A_{1} B_{1} C_{1} D_{1}$ is an isosceles trapezoid, then the isosceles trapezoid $A_{1} B_{1} C_{1} D_{1}$ must have a base parallel to one of the diagonals of the square $A B C D$.
Then, ( ).
(A) (1) and (2) are both true propositions
(B) (1) and (2) are both false propositions
(C) (1) is a false proposition, (2) is a true proposition
(D) (1) is a true proposition, (2) is a false proposition
4. (D).
(1) As shown in the figure, it is easy to know that $\angle A D_{1} A_{1} + \angle D D_{1} C_{1} = 90^{\circ}, \angle D D_{1} C_{1} + \angle D C_{1} D_{1} = 90^{\circ}, \angle A D_{1} A_{1} = \angle D C_{1} D_{1}$, thus $\triangle A D_{1} A_{1} \sim \triangle D C_{1} D_{1}$. Also, Rt $\triangle A D_{1} A_{1} \cong \operatorname{Rt} \triangle C B_{1} C_{1}, A D_{1} = C B_{1}, A A_{1} = C C_{1}$, hence $D D_{1} = B B_{1}, B A_{1} = D C_{1}$. Since $A_{1} D_{1} \neq D_{1} C_{1}$, $\triangle A D_{1} A_{1}$ and $\triangle D C_{1} D_{1}$ are not congruent. Therefore, $A A_{1} \neq D D_{1}$. Without loss of generality, assume $D D_{1} > A A_{1}$. Now assume $A_{1} D_{1} X B D$, then $\angle A A_{1} D_{1} \neq \angle A D_{1} A_{1}$. Without loss of generality, assume $\angle A A_{1} D_{1} > \angle A D_{1} A_{1}$. Thus, it is easy to know that $D D_{1} > D C_{1}$, so $D D_{1} - A D_{1} > |D C_{1} - A A_{1}|$, i.e., $D D_{1} - C B_{1} > |D C_{1} - A A_{1}|$. Draw $D_{1} E \perp B C$ at $E$, $A_{1} F \perp D C$ at $F$, connect $A_{1} C_{1}$, $B_{1} D_{1}$, then $B_{1} E = B B_{1} - A D_{1}, C_{1} F = |D C_{1} - A A_{1}|, A_{1} F = D_{1} E$. We have $B_{1} D_{1} = \sqrt{D_{1} E^{2} + B_{1} E^{2}} > \sqrt{A_{1} F^{2} + C_{1} F^{2}} = A_{1} C_{1}$. This contradicts the fact that the diagonals of a rectangle are equal. Therefore, $A_{1} D_{1} \parallel B D$.
(2) As shown in the figure, through a non-midpoint $P$ on the diagonal $A C$, we can draw $B_{1} D_{1}$ intersecting $B C$ and $A D$ at points $B_{1}$ and $D_{1}$ respectively, such that $\angle A P D_{1} \neq 45^{\circ}$. Draw $A_{1} C_{1} \perp B_{1} D_{1}$ through $P$, intersecting $A B$ and $C D$ at points $A_{1}$ and $C_{1}$ respectively. Therefore, $A, A_{1}, P, D_{1}$ are concyclic, and $C, C_{1}, P, B_{1}$ are concyclic, $\angle A_{1} D_{1} P = \angle A_{1} A P = 45^{\circ} = \angle P A D_{1} = \angle D_{1} A_{1} P$, $\angle P B_{1} C_{1} = \angle P C C_{1} = 45^{\circ} = \angle P C B_{1} = \angle P C_{1} B_{1}$, thus $\angle A_{1} D_{1} P = \angle P B_{1} C_{1}, A_{1} D_{1} \parallel B_{1} C_{1}, A_{1} P = P D_{1}, P C_{1} = B_{1} P, A_{1} C_{1} = B_{1} D_{1}$, and it is easy to know that $A_{1} D_{1} \neq B_{1} C_{1}$. Therefore, $A_{1} B_{1} C_{1} D_{1}$ is an isosceles trapezoid.

Clearly, its bases are not parallel to the diagonals of the square $A B C D$. Counterexamples do exist.