Proving area of triangle formed at parallelogram midpoint is 1/4 of the parallelogram? ABCD is a parallelogram . X is the midpoint of AD & Y is the midpoint of BC. Show that the area of $\triangle {ABX}$ is $\frac{1}{4}$ the area of ABCD

Can you help me with this proof? Where should I start? I think It should be by proving
$\triangle{DBC} \cong \triangle{DBA} $ using SAS as DB is a common side DC= AB as ABCD is a parallelogram, $\angle {BDC} = \angle{DBA} $ alternate angles
And I can also predict that the use of the midpoint theorem here.
Many thanks!


Hint: $Y$ is the midpoint of $\overline{BC}$.
Use this to divide the parallelogram into four triangles and show that these are congruent.
