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(i) It bisects \(\angle{C}\) also,

(ii) ABCD is a rhombus.

Answer :

Given, diagonal AC of a parallelogram ABCD bisects \(\angle{A}\).

\(\therefore \) \(\angle{DAC}\) = \(\angle{BAC}\) = \(\frac{1}{2} \) \(\angle{BAD}\) ...(i)

Here, AB || CD and AC is the transversal.

\(\angle{DCA}\) = \(\angle{CAB}\) ...(ii)(alternate angles)

\(\angle{BCA}\) = \(\angle{DAC}\) ...(iii)(alternate angles)

From eq. (i), (ii) and (iii), we get,

\(\angle{DAC}\) = \(\angle{BAC}\) = \(\angle{DCA}\) = \(\angle{BCA}\)

Now, \(\angle{BCD}\) = \(\angle{BCA}\) + \(\angle{DCA}\)

\(\Rightarrow \) \(\angle{BCD}\)= \(\angle{DAC}\) + \(\angle{CAB}\)

\(\therefore \) \(\angle{BCD}\) = \(\angle{BAD}\)

Thus, we can say that, diagonal AC bisects \(\angle{C}\).

Now, in \(\triangle{OAD}\) and \(\triangle{OCD}\), we have,

OA = OC

(since, diagonals bisects each other)

DO = OD ...(Common side)

\(\angle{AOD}\) = \(\angle{COD}\) = \(90^\circ\)

\(\therefore \) \(\triangle{OAD}\) \(\displaystyle \cong \) \(\triangle{OCD}\)

(By SAS rule)

\(\therefore\) AD = CD ...(By CPCT)

Now, AB = CD and AD = BC

(sides of parallelogram)

\(\therefore \) AB = CD = AD = BC

Hence it is proved that, ABCD is a rhombus.

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