Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
EasyIn the figure,$E$ is any point on the median $AD$ of a $\Delta ABC$. Show that $\text{ar} (ABE) = \text{ar} (ACE)$.
(N/A) We have a $\Delta ABC$ such that $AD$ is a median.
Since a median of a triangle divides it into two triangles of equal areas,
$\text{ar} (\Delta ABD) = \text{ar} (\Delta ADC) \quad \dots(1)$
Similarly,in $\Delta EBC$,$ED$ is a median.
Therefore,$\text{ar} (\Delta EBD) = \text{ar} (\Delta ECD) \quad \dots(2)$
Subtracting equation $(2)$ from equation $(1)$,we get:
$\text{ar} (\Delta ABD) - \text{ar} (\Delta EBD) = \text{ar} (\Delta ADC) - \text{ar} (\Delta ECD)$
$\Rightarrow \text{ar} (\Delta ABE) = \text{ar} (\Delta ACE)$.
Since a median of a triangle divides it into two triangles of equal areas,
$\text{ar} (\Delta ABD) = \text{ar} (\Delta ADC) \quad \dots(1)$
Similarly,in $\Delta EBC$,$ED$ is a median.
Therefore,$\text{ar} (\Delta EBD) = \text{ar} (\Delta ECD) \quad \dots(2)$
Subtracting equation $(2)$ from equation $(1)$,we get:
$\text{ar} (\Delta ABD) - \text{ar} (\Delta EBD) = \text{ar} (\Delta ADC) - \text{ar} (\Delta ECD)$
$\Rightarrow \text{ar} (\Delta ABE) = \text{ar} (\Delta ACE)$.
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