In the figure, $D$ and $E$ are two points on $BC$ such that $BD = DE = EC$. Show that $ar(\triangle ABD) = ar(\triangle ADE) = ar(\triangle AEC)$.
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Given:
$D$ and $E$ are two points on $BC$ such that $BD = DE = EC$.
To do:
We have to show that $ar(\triangle ABD) = ar(\triangle ADE) = ar(\triangle AEC)$.
Solution:
From the figure,
$AL \perp BC$ and $XAY \parallel BC$
$ BD = DE = EC$
$\triangle ABD, \triangle ADE$ and $\triangle AEC$ have equal bases and from the common vertex $A$.
Therefore,
$ar(\triangle ABD) = ar(\triangle ADE) = ar(\triangle AEC)$
Hence proved.
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