In $\Delta ABC$,point $D$ lies on side $BC$. $E$ is the midpoint of $AD$. Prove that,$ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.

(N/A) Given: In $\Delta ABC$,$D$ is a point on $BC$ and $E$ is the midpoint of $AD$.
To prove: $ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.
Proof:
$1$. In $\Delta ABD$,$E$ is the midpoint of $AD$. Since $BE$ is a median of $\Delta ABD$,it divides the triangle into two triangles of equal area. Therefore,$ar(\Delta EBD) = \frac{1}{2} ar(\Delta ABD)$.
$2$. Similarly,in $\Delta ADC$,$E$ is the midpoint of $AD$. Since $CE$ is a median of $\Delta ADC$,it divides the triangle into two triangles of equal area. Therefore,$ar(\Delta ECD) = \frac{1}{2} ar(\Delta ADC)$.
$3$. Adding the two equations: $ar(\Delta EBD) + ar(\Delta ECD) = \frac{1}{2} ar(\Delta ABD) + \frac{1}{2} ar(\Delta ADC)$.
$4$. $ar(\Delta EBC) = \frac{1}{2} [ar(\Delta ABD) + ar(\Delta ADC)]$.
$5$. Since $ar(\Delta ABD) + ar(\Delta ADC) = ar(\Delta ABC)$,we get $ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.
Hence proved.