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

(N/A) Given: $A$ parallelogram $ABCD$.
To Prove: $A$ diagonal,say $AC$,of parallelogram $ABCD$ divides it into congruent triangles $ABC$ and $CDA$. i.e.,$\Delta ABC \cong \Delta CDA$.
Construction: Join $AC$.
Proof:
Since $ABCD$ is a parallelogram,
$AB \parallel CD$ and $AD \parallel BC$.
Now,$AD \parallel BC$ and transversal $AC$ intersects them at $A$ and $C$ respectively.
Therefore,$\angle DAC = \angle BCA$ (Alternate interior angles)........$(1)$
Again,$AB \parallel DC$ and transversal $AC$ intersects them at $A$ and $C$ respectively.
Therefore,$\angle BAC = \angle DCA$ (Alternate interior angles)........$(2)$
Now,in $\Delta ABC$ and $\Delta CDA$,we have:
$\angle BCA = \angle DAC$ [From $(1)$]
$AC = CA$ (Common side)
$\angle BAC = \angle DCA$ [From $(2)$]
So,by $ASA$ congruence criterion,we obtain:
$\Delta ABC \cong \Delta CDA$.