solve this maths.....
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HETY is a parallelogram.
HT and EY are diagonals.
we know that diagonals divides the parallelogram into two equal parts.
so ar(HET) = ar(HTY)
and, ar(HEY) = ar(EYT)
now, in ΔHET, diagonal EY bisects the line segment HT and also the ΔHET,
∴ar(ΔHOE) = ar(ΔEOT)
similarly in ΔETY
ar(ΔEOT) = ar(ΔTOY)
and in ΔHTY,
ar(ΔTOY) = ar(ΔHOY)
That means diagonals in parallelogram divides it into four equal parts.
Hence proved .
HT and EY are diagonals.
we know that diagonals divides the parallelogram into two equal parts.
so ar(HET) = ar(HTY)
and, ar(HEY) = ar(EYT)
now, in ΔHET, diagonal EY bisects the line segment HT and also the ΔHET,
∴ar(ΔHOE) = ar(ΔEOT)
similarly in ΔETY
ar(ΔEOT) = ar(ΔTOY)
and in ΔHTY,
ar(ΔTOY) = ar(ΔHOY)
That means diagonals in parallelogram divides it into four equal parts.
Hence proved .
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