Question

In a triangle $ABC,E$ is the midpoint of median AD, show that $area△ABE=\frac{1}{4}area\left(△ABC\right)$

Answer 1

In given figure AD is the median of $\Delta ABC$.

Therefore, it will divide $\Delta ABC$ into two triangles of equal area.

Area $\Delta ABD$ = Area $\Delta ACD$

$\therefore area\left(\Delta ABD\right)=\frac{1}{2}area\left(\Delta ABC\right)$ ------------(1)

In $\Delta ABD$, E is the mid-point of AD.

Therefore, BE is the median.

Area $\Delta ABE$ = Area $\Delta BED$

$\therefore area\left(\Delta ABE\right)=\frac{1}{2}area\left(\Delta ABD\right)$ ------------(2)

From (1) and (2) we get

$area\left(\Delta ABE\right)=\frac{1}{2}\times \frac{1}{2}area\left(\Delta ABC\right)$

$area\left(\Delta ABE\right)=\frac{1}{4}area\left(\Delta ABC\right)$ $\left[henceproved\right]$