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

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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.

Updated on 10-Oct-2022 13:41:49