Let us join AC.

In ΔABC,

BC = AB (Sides of a rhombus are equal to each other)

∠1 = ∠2 (Angles opposite to equal sides of a triangle are equal)

However,

∠1 = ∠3 (Alternate interior angles for parallel lines AB and CD)

∠2 = ∠3

Therefore, AC bisects C

Also,

∠2 = ∠4 (Alternate interior angles for || lines BC and DA)

∠1 = ∠4

Therefore,

AC bisects A

Similarly, it can be proved that BD bisects B and D as well