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The diagonals of a parallelogram A B ...

The diagonals of a parallelogram `A B C D` intersect at `Odot` A line through `O` meets `A B\ ` in `x\ a n d\ C D` in `Ydot` Show that `a r\ (A X Y X)=1/2(a r|""|^(gm)\ \ A B C D)`

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We know that, a diagonal of a parallelogram divides it into two triangles of equal areas.
Now, AC is the diagonal of parallelogram ABCD
`:. " area of "DeltaACD = (1)/(2) xx " are of " square ABCD`...(1)
In `Delta AOX and Delta COY`,
`.:{(AO = CO),(angleAOX = angleCOY),(angleOAX = angleOCY):}`
`:. Delta AOX ~= Delta COY`
`rArr` area of `Delta AOX =` area of `DeltaCOY`
Now, area of `Delta ACD =` are of `squareAOYD + " area of " Delta OCY`
`- " area of " squareAOYD + " area of " Delta AOX` [from (2)]
= area of `squareAXYD`
`:.` From (1), we get
area of `square AXYD = (1)/(2) xx " area of " square ABCD`
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