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# 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,
$$\therefore$$ $$\triangle{AOD}$$ $$\displaystyle \cong$$ $$\triangle{COD}$$ ...(By SSS rule)
$$\therefore$$ $$\angle{ADO}$$ = $$\angle{CDO}$$ ...(By CPCT)
Hence, it is proved that, Diagonal BD bisects $$\angle{D}$$, too.