$A$ diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.

(N/A) $ABCD$ is a parallelogram and diagonal $AC$ bisects $\angle A$. We have to show that $ABCD$ is a rhombus.
$\angle 1 = \angle 2$ $\dots(1)$ [$\because AC$ bisects $\angle A$]
$\angle 2 = \angle 4$ $\dots(2)$ [Alternate interior angles,since $AB \parallel DC$]
From $(1)$ and $(2)$,we get:
$\angle 1 = \angle 4$
Now,in $\triangle ABC$,we have:
$\angle 1 = \angle 4$ [Proved above]
$BC = AB$ [$\because$ Sides opposite to equal angles are equal]
Also,$AB = DC$ and $AD = BC$ [$\because$ Opposite sides of a parallelogram are equal]
Therefore,$AB = BC = CD = AD$.
Since all sides of the parallelogram $ABCD$ are equal,it is a rhombus.