Diagonal $AC$ of a parallelogram $ABCD$ bisects $\angle A$ (see Fig). Show that it bisects $\angle C$ also.

(N/A) Given: $ABCD$ is a parallelogram in which diagonal $AC$ bisects $\angle A$. Thus,$\angle DAC = \angle BAC$.
To prove: $AC$ bisects $\angle C$,i.e.,$\angle DCA = \angle BCA$.
Proof:
$1$. Since $ABCD$ is a parallelogram,$AB \parallel DC$ and $AC$ is a transversal.
Therefore,$\angle BAC = \angle DCA$ (Alternate interior angles) ....... $(1)$
$2$. Also,$BC \parallel AD$ and $AC$ is a transversal.
Therefore,$\angle DAC = \angle BCA$ (Alternate interior angles) ....... $(2)$
$3$. Since $AC$ bisects $\angle A$,we have $\angle DAC = \angle BAC$ ....... $(3)$
$4$. From equations $(1)$,$(2)$,and $(3)$,we get:
$\angle DCA = \angle BCA$
Hence,$AC$ bisects $\angle C$.