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!


Observe $$ar(BAXY)=ar(DXYC)$$

Since $BAXY$ and $DXYC$ are also parallelograms and we know that the diagonal of a parallelogram divides it into two triangles of equal area.
$$ar(ABCD)=2\times\frac 12 ar(ABCD)$$ $$ar(ABCD)=ar(BAXY)+ar(DXYC)$$ $$ar(ABCD)=2\times \frac{1}{2}[ar(BAXY)+ar(DXYC)]$$ $$ar(ABCD)=2\times [ar(\triangle ABX)+ar(\triangle BXY)+ar(\triangle XYD)+ar(\triangle YDC)]$$ $$ar(ABCD)=4\times ar(\triangle ABX)=4\times ar(\triangle BXY)=4\times ar(\triangle XYD)=4\times ar(\triangle YDC)$$
