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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)`

Text Solution

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Diagonal AC of ||gm ABCD divides it into two triangles of equal area.
`therefore ar(squareACD)=(1)/(2)ar("||"gm ABCD)" "...(i)`
In `triangle` OAP and OCQ, we have:
OA = OC (diagonals of a ||gm bisect each other)
`angleAOP=angleCOQ " "("vert. opp." angle)`
`angle PAO=angleQCO " " ("alt. interior " angle)`
`therefore triangle OAPcongtriangleOCQ.`
`therefore ar(triangleOAP)=ar(triangleOCQ)`
`rArr ar(triangleOAP)+ar("quad. AOQD")`
`=ar(triangleOCQ)+ar("quad. AOQD")`
`rArr ar("quad. APQD")=ar(triangleACD)`
`=(1)/(2)ar("||gm ABCD") " " ["using (i)"]`.
`therefore ar(squareAPQD)=(1)/(2)ar("||gm ABCD")`.
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