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# In figure below, $E$ is any point on median $ \mathrm{AD} $ of a $ \triangle \mathrm{ABC} $. Show that ar $ (\mathrm{ABE})=\operatorname{ar}(\mathrm{ACE}) $.

"

Given:

$E$ is any point on median \( \mathrm{AD} \) of a \( \triangle \mathrm{ABC} \).

To do:

We have to show that ar \( (\mathrm{ABE})=\operatorname{ar}(\mathrm{ACE}) \).

Solution:

$AD$ is the median of $\triangle ABC$.

This implies,

$AD$ divides $\triangle ABC$ into two triangles of equal area.

Therefore,

$ar(\triangle ABD) = ar(\triangle ACD)$...........(i)

$AD$ is the median of $\triangle ABC$.

This implies,

$ED$ is the median of $\triangle ABC$

Therefore,

$\triangle ar(EBD) = ar(\triangle ECD)$.........(ii)

Subtracting (ii) from (i), we get,

$ar(\triangle ABD) - ar(\triangle EBD) = ar(\triangle ACD) - ar(\triangle ECD)$

$ar(\triangle ABE) = ar(\triangle ACE)$

Hence proved.

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