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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
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