# 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 ? 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 :)

• The point $Y$ is there for a reason - draw XY and YD and observe that you get four congruent triangles! – Parcly Taxel Oct 19 '16 at 4:15
• ok i updated the diagram , now what would be the case for congruency ? – user366082 Oct 19 '16 at 4:20
• The problem is solved by the diagram alone. This is called a proof without words and they crop up often in geometry. – Parcly Taxel Oct 19 '16 at 4:21
• @user366082 All corresponding angles and sides having the same magnitude, due to properties of parallelograms. – Graham Kemp Oct 19 '16 at 4:25

Hint: $Y$ is the midpoint of $\overline{BC}$.

Use this to divide the parallelogram into four triangles and show that these are congruent.

• Using what condition? – user366082 Oct 19 '16 at 4:18
• @user366082 It's a parallelogram. – Graham Kemp Oct 19 '16 at 4:22
• what case like side angle side should be uesd? – user366082 Oct 19 '16 at 6:28

Length of perpendicular for the triangle and parallelogram is same.

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

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