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(i) ABCD is square

(ii) diagonal BD bisects $\angle $${\text{B}}$ as well as $\angle $${\text{D}}$.

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(i) we need to prove ABCD is square.

Given in question,

ABCD is a rectangle, in which diagonal AC bisect $\angle $${\text{A}}$ as well as $\angle $${\text{C}}$.

Therefore,

$\angle $${\text{DAC}}$=$\angle $${\text{CAB}}$……………… (1)

$\angle $${\text{DCA}}$=$\angle $${\text{BCA}}$………………. (2)

A square is a rectangle when all sides are equal. Now,

AD$\parallel $BC & AC is transversal, therefore

$\angle$${\text{DAC}}$=$\angle$${\text{BCA}}$ [Alternate angles]

From (1),

$\angle $${\text{CAB}}$=$\angle $${\text{BCA}}$……………….. (3)

In $\vartriangle $ABC,

$\angle $${\text{CAB}}$=$\angle $${\text{BCA}}$, therefore

BC=AB………………….. (4) [sides opposite to equal angles]

But BC=AD & AB=DC………………….. (5) [Opposite sides of rectangle]

Therefore from (4) & (5),

AB=BC=CD=AD

(ii) we need to prove that diagonal BD bisects $\angle $${\text{B}}$ as well as $\angle $${\text{D}}$.

Now, As we discussed in the previous part.

ABCD is a square and we know that diagonals of a square bisect its angles.