How to prove that diagonals of rhombus bisects its angles?

Given: ABCD is a Rhombus

 ⇒AB=BC=CD=AD

To Prove:

∠BAC=∠DAC

and

∠DCA=∠BCA

Proof: 

In ΔADC and  ΔABC

AB=AD  [Given]

CD=CB  [Given]

AC=AC  [Common]

⇒ΔADCΔABC[By SSS congruence criterion]

∠BAC=∠DAC [By cpct]

∠DCA=∠BCA  [By cpct]

Hence the result.