ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Which of the following statements is true?

In a rectangle ABCD, diagonal AC bisects ∠A and ∠C.
Let's consider the triangle ABC. Since ABCD is a rectangle, AB is parallel to CD and BC is parallel to AD. Also, ∠A = ∠B = ∠C = ∠D = 90°. Given that AC bisects ∠A and ∠C, we have ∠BAC = ∠DAC and ∠BCA = ∠DCA.
In ΔABC, ∠BAC + ∠BCA + ∠ABC = 180°
∠BAC + ∠BCA + 90° = 180°
∠BAC + ∠BCA = 90°
Since ∠BAC = ∠DAC and ∠BCA = ∠DCA, we have
2∠BAC + 2∠BCA = 180°
∠BAC + ∠BCA = 45°
This is a contradiction to our earlier result that ∠BAC + ∠BCA = 90°.
However, in a rectangle, the diagonals bisect each other. Also, the diagonals are equal in length. Let's assume ABCD is a rectangle and AC bisects ∠A and ∠C. Since AC is a diagonal, it bisects the angles at A and C only if ABCD is a square. If ABCD is a square, then AB=BC=CD=DA and AC bisects ∠A and ∠C. This means that ΔABC is an isosceles right-angled triangle. Then, ∠BAC = ∠BCA = 45°. Similarly, ΔADC is also an isosceles right-angled triangle. Then ∠DAC = ∠DCA = 45°. Therefore, if AC bisects ∠A and ∠C in rectangle ABCD, then ABCD must be a square.