Question

In the given figure, D is the mid-point of BC and E is any point on AD. Prove that :
(i) ar $\left(△EBD\right)$ $=$ ar $\left(△EDC\right).$
(ii) ar $\left(△ABE\right)$ $=$ ar $\left(△ACE\right).$

Answer 1

In $△ABC$

D is mid point of $BC$

$\Rightarrow AD$ is the median.

similarly In $△EBC$

D is mid point of BC

$\Rightarrow ED$ is median.

As median divides a given triangle

in two triangles of equal area

$\Rightarrow ar\left(△ABD\right)=ar\left(△ACD\right)$ ___(1)

$ar\left(△EBD\right)=ar\left(△ECD\right)$ __(2)

(1)-(2) statement

$ar\left(△ABD\right)-ar\left(△EBD\right)=ar\left(△ACD\right)-ar\left(△ECD\right)$

$\Rightarrow ar\left(△ABE\right)=ar\left(△ACE\right)$__(3)

statement (2) and (3) are proof of

part (I) and (II) respectively.