$ABCD$ is a rhombus. Show that diagonal $AC$ bisects $\angle A$ as well as $\angle C$ and diagonal $BD$ bisects $\angle B$ as well as $\angle D$.

(N/A) $ABCD$ is a rhombus.
$AB = BC = CD = AD$
Also,$AB \parallel CD$ and $AD \parallel BC$.
Now,in $\triangle ADC$,$AD = CD$ (sides of a rhombus).
Therefore,$\angle 1 = \angle 2$ (angles opposite to equal sides are equal) ....... $(1)$
Also,$CD \parallel AB$ (opposite sides of a parallelogram) and $AC$ is a transversal.
Therefore,$\angle 2 = \angle 3$ (alternate interior angles) ....... $(2)$
From $(1)$ and $(2)$,we have $\angle 1 = \angle 3$.
Similarly,since $AD \parallel BC$ and $AC$ is a transversal,$\angle 1 = \angle 4$ (alternate interior angles).
Since $\angle 1 = \angle 2$ and $\angle 1 = \angle 4$,we have $\angle 2 = \angle 4$ is not necessarily true,but rather $\angle 1 = \angle 4$ and $\angle 2 = \angle 3$ implies $AC$ bisects $\angle A$ and $\angle C$.
Thus,$AC$ bisects $\angle A$ and $\angle C$.
Similarly,we can prove that $BD$ bisects $\angle B$ and $\angle D$.