Question

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​Show that the diagonals of a parallelogram divide it into four triangles of equal area.

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Answer 1

we know that diagonal of a parallelogram

bisect each other

there fore AO is equal to OC

and

DO is equal to OB

in parallelogram ABCD

AC is the diagonal and O is the median of triangle divide it into two eqaul parts of

equal triangles

this implies in triangle ABC, oc is median there fore

ar(AOC)is equal to ar (BOC) (1)

similarly in triangle CBD, OB is median ar(COB) is equal to ar(BOD) (2)

in trianle BAD, OD is median ar(BOD) is equal to

ar(AOD) (3)

NOW

from 1,2 and 3 we get

ar(AOC)=ar(BOC)=ar(BOD)=ar(AOD) hence, prooved

Answer 2

$\huge\purple{\boxed{\green{\mathbb{\overbrace{\underbrace{\fcolorbox{purple}{aqua}{\underline{\red{Answer : -}}}}}}}}}$

we know that diagonal of a parallelogram bisect each other

there fore AO is equal to OC

and

DO is equal to OB

in parallelogram ABCD

AC is the diagonal and O is the median of triangle divide it into two eqaul parts of equal triangles

this implies in triangle ABC, oc is median there fore

ar(AOC)is equal to ar (BOC) (1)

similarly in triangle CBD, OB is median

ar(COB) is equal to ar(BOD) (2)

in trianle BAD, OD is median

ar(BOD) is equal to ar(AOD) (3)

NOW

from 1,2 and 3 we get

ar(AOC)=ar(BOC)=ar(BOD)=ar(AOD)

hence, prooved

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Hope it helps