MHB Proof of Parallelogram ABCD: Midpoint X & Y Show Area $\frac{1}{4}$

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To prove that the area of triangle ABX is one-fourth the area of parallelogram ABCD, start by establishing that triangles DBC and DBA are congruent using the SAS criterion. Next, construct segment XY, connecting midpoints X and Y of sides AD and BC, respectively, which leads to the conclusion that triangles ABX and XYB are congruent by SSS. The midpoint theorem indicates that AX equals XD and BY equals YC, reinforcing the relationships between the segments. Additionally, drawing a perpendicular from point X to line AB shows that its length is half that of the perpendicular from point D to line AB, allowing for the area calculations. Thus, the area of triangle ABX can be confirmed as one-fourth that of parallelogram ABCD.
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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 :)
 

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Start by constructing segment XY. What can be said about triangles ABX and XYB?
 
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.
 
greg1313 said:
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 :)
 

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I'd use SSS.
 
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 :)
 
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