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.
