ABCD is a rectangle in which diagonal AC bisects \[\angle {\text{A}}\] as well as \[\angle {\text{C}}\]. Show that
$(i)$ ABCD is a square
$(ii)$ diagonal BD bisects \[\angle {\text{B}}\] as well as \[\angle {\text{D}}\].

Hint: Here, we will proceed by using the concept of alternative angles with the help of which we will prove some angles in a triangle equal and then we will use the concept that in any triangle if two angles are equal then the sides opposite to those angles will also be equal in lengths.

Complete Step-by-Step solution:
$(i)$ To prove: ABCD is a square
Let us consider ABCD as a rectangle in which diagonal AC bisects \[\angle {\text{A}}\] as well as \[\angle {\text{C}}\]
i.e., \[\angle {\text{OAB}} = \angle {\text{OAD }} \to {\text{(1)}}\] and \[\angle {\text{OCB}} = \angle {\text{OCD }} \to {\text{(2)}}\]
Since, ABCD is a rectangle so the pairs of opposite sides will be equal and parallel.
i.e., AB = CD and AB$\parallel $CD, BC = AD and BC$\parallel $AD
Since, \[\angle {\text{OAD}}\] and \[\angle {\text{OCB}}\] are alternate angles with AC as transverse and BC$\parallel $AD.
As we know that any two alternative angles are equal in measure.
Then, \[\angle {\text{OAD}} = \angle {\text{OCB }} \to {\text{(3)}}\]
By comparing equations (1) and (3), we get
\[\angle {\text{OAB}} = \angle {\text{OCB}}\]
Also we know that in any triangle if two angles are equal then the sides opposite to those angles will also be equal in lengths.
Consider triangle ABC, we have
\[\angle {\text{OAB}} = \angle {\text{OCB}}\]
The side opposite to \[\angle {\text{OAB}}\] is BC and the side opposite to \[\angle {\text{OCB}}\] is AB
Therefore, BC = AB
Since, AB = CD and BC = AB and BC = AD
So, AB = BC = CD = AD i.e., all the sides of the given rectangle ABCD are equal in lengths.
According to the property of any square, all the sides of the square should always be equal in lengths
Therefore, ABCD is a square.
$(ii)$ To prove: Diagonal BD bisects \[\angle {\text{B}}\] as well as \[\angle {\text{D}}\] i.e., \[\angle {\text{OBA}} = \angle {\text{OBC}}\] and \[\angle {\text{ODA}} = \angle {\text{ODC}}\]
According to the properties of any square, the diagonals of that square bisects the interior angles made at the vertices.
Since, ABCD is a square which means the diagonal AC bisects \[\angle {\text{A}}\] as well as \[\angle {\text{C}}\] and the diagonal BD bisects \[\angle {\text{B}}\] as well as \[\angle {\text{D}}\].
Therefore, the diagonal BD bisects \[\angle {\text{B}}\] as well as \[\angle {\text{D}}\] i.e., \[\angle {\text{OBA}} = \angle {\text{OBC}}\] and \[\angle {\text{ODA}} = \angle {\text{ODC}}\].

Note: A square is a quadrilateral with all four angles right angles and all four sides of the same length. So a square is a special kind of rectangle. In this particular problem, in order to prove the second part, we have directly used the property of square. Here we can also prove that by using the inverse of the concept used in proving ABCD is a rectangle.