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)`

Explanation

As we know that, a diagonal of a parallelogram divides it into two triangles of equal areas.

Now, AC is the diagonal of parallelogram ABCD.

`therefore` Area of `triangleACD`=`1/2` area of square ABCD_______1

In `triangleAOX` and `triangleCOY`

`AO=CO`

`angleAOX=angleCOY`

`angleOAX=angleOCY`

`therefore` `triangleAOX~=triangleCOY`

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