ABCD is a rectangle in which diagonal AC bisects $\angle A$ as well as $\angle C$.
Show that:
i) ABCD is a square.
ii) Diagonal BD bisects $\angle B$ as well as$\angle D$.

Hint: Rectangle: a quadrilateral in which opposite sides are parallel and equal and all of the angles are ${90^ \circ }$ Square: A square is a quadrilateral in which all four sides are equal where both pairs of opposite sides are parallel and all angles are ${90^ \circ }$.

Complete step-by-step answer:

Proof: In $\Delta ABC$ and $\Delta ADC$
                                  $\angle BAC = \angle DAC$ ($AC$ bisects $\angle A$)
                                   $\angle BCA = \angle DCA$ ($AC$ bisects $\angle C$)
                                          $AC = AC$ (common)
Therefore by Angle-Side-Angle Congruence, the triangles are congruent.
Therefore, by using congruent parts of congruent triangles (CPCT), we can say that,
$AB = AD$
And,$CB = CD$.
But, we know that, In a rectangle opposite sides are equal,
Therefore,
$AB = DC$ and $BC = AD$
Therefore by taking all the conditions proved above we can say that,
$AB = BC = CD = AD$
Since, all the four sides are proved to be equal, therefore, we can say that it is a square.
Hence Proved!
Proof: In $\Delta ABD$ and $\Delta CDB$
                         $AD = CB$(Equal sides of square)
                          $AB = CD$ (Equal sides of square)
                          $BD = BD$ (Common)
Therefore by side-side-side congruence, the triangles are congruent.
Therefore, by using congruent parts of congruent triangles (CPCT), we can say that,
$\angle ABD = \angle CBD$ and,
$\angle ADB = \angle CDB$
Therefore, we can say that diagonal $BD$ bisects $\angle B$ as well as$\angle D$.

Note: Make sure you write the reason in the bracket for the statements you write while proving the congruence.