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# 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.