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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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