Diagonal AC of a parallelogram ABCD bisects \(\angle{A}\) (see figure). Show that
(i) It bisects \(\angle{C}\) also,
(ii) ABCD is a rhombus.

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.