Given : A parallelogram ABCD, in which diagonal AC bisects ∠A, i.e., ∠DAC = ∠BAC.

**To Prove : **

(i) Diagonal AC bisects

∠C i.e., ∠DCA = ∠BCA

(ii) ABCD is a rhomhus.

**Proof :**

(i) ∠DAC = ∠BCA [Alternate angles]

∠BAC = ∠DCA [Alternate angles]

But, ∠DAC = ∠BAC [Given]

∴ ∠BCA = ∠DCA

Hence, AC bisects ∠DCB

Or, AC bisects ∠C **Proved. **

(ii) In ∆ABC and ∆CDA

AC = AC [Common]

∠BAC = ∠DAC [Given]

and ∠BCA = ∠DAC [Proved above]

∴ ∆ABC ≅ ∆ADC [ASA congruence]

∴ BC = DC [CPCT]

But AB = DC [Given]

∴ AB = BC = DC = AD

Hence, ABCD is a rhombus **Proved.**