Prove that a diagonal of a parallelogram divides it into two congruent triangles.

(N/A) Let $ABCD$ be a parallelogram and $AC$ be a diagonal. Observe that the diagonal $AC$ divides parallelogram $ABCD$ into two triangles,namely,$\Delta ABC$ and $\Delta CDA$. We need to prove that these triangles are congruent.
In $\Delta ABC$ and $\Delta CDA$,note that $BC \parallel AD$ and $AC$ is a transversal.
So,$\angle BCA = \angle DAC$ (Pair of alternate interior angles).
Also,$AB \parallel DC$ and $AC$ is a transversal.
So,$\angle BAC = \angle DCA$ (Pair of alternate interior angles).
And $AC = CA$ (Common side).
Therefore,$\Delta ABC \cong \Delta CDA$ (by $ASA$ congruence rule).
Thus,the diagonal $AC$ divides the parallelogram $ABCD$ into two congruent triangles $ABC$ and $CDA$.