Question

If D, E and F are the mid points of sides BC, CA and AB respectively of a $\Delta ABC$, then using coordinate geometry, prove that

Area of $\Delta DEF=\frac{1}{4}\left(Areaof\Delta ABC\right)$

Answer 1

Let $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ be the vertices of $\Delta ABC$. Then, the coordinates of D, E and F are $(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2})(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2})and(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ respectively.

${\Delta }_1=Areaof\Delta ABC=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$
${\Delta }_2=Areaof\Delta DEF=\frac{1}{2}\left(\frac{x_2+x_3}{2}\right)(\frac{y_2+y_3}{2}-\frac{y_1+y_2}{2})+\left(\frac{x_1+x_3}{2}\right)(\frac{y_1+y_2}{2}-\frac{y_2+y_3}{2})+\left(\frac{x_1+x_2}{2}\right)(\frac{y_2+y_3}{2}-\frac{y_1+y_3}{2})$
${\Delta }_2=\frac{1}{8}\left|\right(x2+x3\left)\right(y3-y2)+(x1+x3\left)\right(y1-y3)+(x1+x2\left)\right(y2-y1\left)\right|$
${\Delta }_2=\frac{1}{8}\left|x1\right(y2-y3)+x2(y3-y1)+x3(y1-y2\left)\right|$
${\Delta }_2=\frac{1}{4}\left(Areaof\Delta ABC\right)=\frac{1}{4}{\Delta }_1$