** Given :** ABCD is a rhombus, i.e.,

AB = BC = CD = DA.

**To Prove :** ∠DAC = ∠BAC,

∠BCA = ∠DCA

∠ADB = ∠CDB, ∠ABD = ∠CBD

**Proof :** In ∆ABC and ∆CDA, we have

AB = AD [Sides of a rhombus]

AC = AC [Common]

BC = CD [Sides of a rhombus]

∆ABC ≅ ∆ADC [SSS congruence]

So, ∠DAC = ∠BAC

∠BCA = ∠DCA

Similarly, ∠ADB = ∠CDB and ∠ABD = ∠CBD.

Hence, diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D. **Proved.**