Question

Any point $D$ is taken in the base of a triangle $ABC$ and $AD$ is produced to $E$, making $DE=AD$. Show that $\mathrm{ar}\left(\Delta BCE\right)=\mathrm{ar}\left(\Delta ABC\right)$

Answer 1

$\begin{array}{c}Toprove:\\ Areaof\Delta BCE=areaof\Delta ABC\\ proof:\\ In\Delta ABE,\\ AD=DE\left(given\right)\\ \therefore BDisamedianof\Delta ABE\\ \Rightarrow ar\left(\Delta ABD\right)=ar\left(\Delta BED\right).......\left(1\right)\\ mediandividesatriangleintotwotrianglesofequalarea.\\ Similarly,\\ In\Delta ACE\\ CDismedianof\Delta CED......\left(2\right)\\ Adding\left(1\right)and\left(2\right),\\ ar\left(\Delta ABD\right)ar\left(\Delta ACD\right)=ar\left(\Delta BED\right)+ar\left(\Delta CED\right)\\ orar\left(\Delta ABC\right)=ar\left(\Delta BCE\right)\\ Hence,areaof\Delta BCE=areaof\Delta ABC.\end{array}$