Diagonal $AC$ of a parallelogram $ABCD$ bisects $\angle A$ (see Fig). Show that $ABCD$ is a rhombus.

(N/A) Given: $A$ parallelogram $ABCD$ in which diagonal $AC$ bisects $\angle A$. So,$\angle DAC = \angle BAC$.
To prove: $ABCD$ is a rhombus.
Proof:
$1$. Since $ABCD$ is a parallelogram,$AD \parallel BC$ and $AC$ is a transversal.
Therefore,$\angle DAC = \angle BCA$ (Alternate interior angles).
$2$. We are given that $\angle DAC = \angle BAC$.
$3$. From steps $1$ and $2$,we get $\angle BAC = \angle BCA$.
$4$. In $\Delta ABC$,since $\angle BAC = \angle BCA$,the sides opposite to these angles must be equal.
Therefore,$BC = AB$ (Sides opposite to equal angles are equal).
$5$. In a parallelogram,opposite sides are equal,so $AB = CD$ and $AD = BC$.
$6$. Since $AB = BC$ and $AB = CD, BC = AD$,we have $AB = BC = CD = DA$.
Since all sides of the parallelogram $ABCD$ are equal,it is a rhombus.