Question

In $E$ is any point on median $AD$ of a
$\Delta$ $ABC$. Show that ar $\left(ABE\right)=\mathrm{ar}\left(ACE\right)$

Answer 1

We know, medium divides a triangle into two triangles of equal area

$\therefore$ ar ${\Delta }^{\prime \prime }ABD=ar{\Delta }^{\prime \prime }ACD⟶\left(i\right)$

also, ar ${\Delta }^{\prime \prime }BED=ar{\Delta }^{\prime \prime }CED⟶\left(ii\right)$

on $\left(i\right)-\left(ii\right)$

ar ${\Delta }^{\prime \prime }ABD-ar{\Delta }^{\prime \prime }BED=ar{\Delta }^{\prime \prime }ACD-ar{\Delta }^{\prime \prime }CED$

$\therefore$ ar ${\Delta }^{\prime \prime }ABE=ar{\Delta }^{\prime \prime }ACE$ Proved.