$A$ diagonal of a parallelogram bisects one of its angles. Prove that it will bisect its opposite angle also.

(N/A) Let us consider a parallelogram $ABCD$ where $AC$ is a diagonal that bisects $\angle BAD$.
Given: $\angle BAC = \angle DAC$.
To prove: $\angle BCA = \angle DCA$.
Proof:
Since $AB \parallel CD$ and $AC$ is a transversal,the alternate interior angles are equal:
$\angle BAC = \angle DCA$ --- $(1)$
Since $AD \parallel BC$ and $AC$ is a transversal,the alternate interior angles are equal:
$\angle DAC = \angle BCA$ --- $(2)$
From the given condition,we know:
$\angle BAC = \angle DAC$ --- $(3)$
Comparing equations $(1)$,$(2)$,and $(3)$,we get:
$\angle BCA = \angle DCA$.
Thus,the diagonal $AC$ bisects the opposite angle $\angle BCD$ as well.