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Thediagonals of a parallelogram `A B C D`intersectat `Odot`A linethrough `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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Answer» 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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