The diagonal of a parallelogram divides it into two congruent triangles. 

Concept: A parallelogram's diagonals cross each other. 

Let us consider a parallelogram $ABCD$ with  $AC$as its diagonal.

To prove:  $∆ABC\cong ∆ADC$

Proof:

We know that the opposite sides of a parallelogram are parallel.

Hence  $AB\parallel DC$ and $AD\parallel BC , AC$

is the transversal of the parallel lines $AB$and $DC$$\angle BAC=\angle DCA.......\left(1\right)$

 (Alternate angles) $AC$

is the transversal of the parallel lines AD and  BC  

$\angle DAC=\angle BCA .....\left(2\right)$

In $∆ABC$ and  $∆ADC$  

$\angle BAC=\angle DCA$ , $AC$  is the common side

$$\begin{gathered} \angle DAC=\angle BCA \\ Therefore, \\ ∆ABC\cong ∆ADC \end{gathered}$$