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(i) ABCD is a square

(ii) Diagonal BD bisects \(\angle{B}\) as well as \(\angle{D}\).

Answer :

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.

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