Question

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

Answer 1

Question :-

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

Answer :-

Given :

$\colorbox{black}{\colorbox{red}{\colorbox{white}{ABCD\:is\:a\: parallelograml}}}$

To show :

• ar (∆ABO) = ar (∆BCO)

= ar (∆CDO)

= ar (∆DAO)

Proof :

$\rightarrow$ In parallelogram ABCD

$\rightarrow$ Diagonals AC and BD of bisect each other

$\therefore$ OA = OC and OB = OD

$\rightarrow$ [ O is the midpoint of AC and BD ]

Now ,

$\\tt\:\sf\small\implies$$\colorbox{black}{\colorbox{white}{\colorbox{white}{In \: ∆ABC}}}$

$\because$ OA = OC , and BO is the median of ∆ABC,

So,

$\tt\:\sf\small\colorbox{black}{\colorbox{Black}{\colorbox{white}{ar\:(∆ABO)\:=\:ar\:(∆BCO)}}}$

$\rightarrow$ Median divides ∆ into two equal area ...(i)

Similarly,

$\sf\huge\implies$ $\tt\:ar (∆ADO) = ar (∆CDO)...(ii)$

$\sf\huge\implies$ $\tt\:ar (∆ADO) = ar (∆ABO)....(iii)$

$\sf\huge\implies$$\tt\: ar (∆ADO) = ar (∆CDO)...(iv)<strong> </strong>$

From There ,

$\sf\huge\implies$ ar (∆ABO) = ar (∆BCO)

$\sf\huge\implies$ = ar(∆CDO)

$\sf\huge\implies$ = ar (∆DAO)

$\sf\huge\implies$ Hence proved

Therefore,

$\tt\: ar (∆ABO) = ar (∆BCO) = ar (∆CDO) = ar (∆DAO )$

Hence , diagonals of a parallelogram into four congruent (equal area) triangles

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