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 \(\angle{D}\).

Given: ABCD is a rectangle. AB = CD and BC = AD ...(i)


To prove: (i) ABCD is a square. i.e., AB = BC = CD = DA
(ii) Diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).

Proof: In \(\triangle{ADC}\) and \(\triangle{ABC}\), we have,
\(\angle{DAC}\) = \(\angle{BAC}\)
(Since, AB || DC and AC is transversal that intersects)
Similarly, \(\angle{DCA}\) = \(\angle{BCA}\)
AC = CA ...(Common side)
\(\therefore \) \(\triangle{ADC}\) \(\displaystyle \cong \) \(\triangle{ABC}\) ...(By ASA rule)
\(\therefore \)AD = AB ...(By CPCT)
Also, CD = BC ...(ii)
Thus, from eq. (i) and(ii), we get,
AB = BC = AD = CD

Hence, it is proved that ABCD is a square.

Now, in \(\triangle{AOB}\) and \(\triangle{COB}\), we have,
AB = BC ...(Given)
BO = OB ...(Common side)
OA = OC
(Since,diagonal bisectS each other)
\(\therefore \) \(\triangle{AOB}\) \(\displaystyle \cong \) \(\triangle{COB}\) ...(By SSS rule)
\(\therefore \) \(\angle{OBA}\) = \(\angle{OBC}\) ...(By CPCT)
This shows that, Diagonal BD bisects \(\angle{B}\).

Similarly, Now, in \(\triangle{AOD}\) and \(\triangle{COD}\), we have,
AD = CD ...(Given)
OD = DO ...(Common side)
OA = OC
(Since,diagonal bisectS each other)
\(\therefore \) \(\triangle{AOD}\) \(\displaystyle \cong \) \(\triangle{COD}\) ...(By SSS rule)
\(\therefore \) \(\angle{ADO}\) = \(\angle{CDO}\) ...(By CPCT)

Hence, it is proved that, Diagonal BD bisects \(\angle{D}\), too.