Show that triangle ABX is 1/4 the area of parallelogram ABCD

[mathlearn]

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

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Can you help me with this proof ? were 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 midpoint theorem here

Many Thanks :)

 

[Greg]

Start by constructing segment XY. What can be said about triangles ABX and XYB?

 

[HallsofIvy]

Another way: Draw the perpendicular to AB from X. Show, using "similar triangles", that its length is half the length of the perpendicular to AB from D. The result follows immediately from the formulas for the areas of triangle and parallelogram.

 

[mathlearn]

Start by constructing segment XY. What can be said about triangles ABX and XYB?

They are congruent! But how can they be proved using $SSS$ or $SAS$ or $AAS$

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Many Thanks :)

 

[Greg]

I'd use SSS.

 

[mathlearn]

I wonder what are the uses of joing the two midpoints of the sides x & y.

It helps to state that,

AX=XD
& BY=YC

And can AB=XY be or DC=XY be said using the midpoint theorem?

Many Thanks :)