In the figure, $D$ and $E$ are two points on $BC$ such that $BD = DE = EC$. Show that $ar(\triangle ABD) = ar(\triangle ADE) = ar(\triangle AEC)$.
"

Given:

$D$ and $E$ are two points on $BC$ such that $BD = DE = EC$.

To do:

We have to show that $ar(\triangle ABD) = ar(\triangle ADE) = ar(\triangle AEC)$.

Solution:

From the figure,

$AL \perp BC$ and $XAY \parallel BC$

$ BD = DE = EC$

$\triangle ABD, \triangle ADE$ and $\triangle AEC$ have equal bases and from the common vertex $A$.

Therefore,

$ar(\triangle ABD) = ar(\triangle ADE) = ar(\triangle AEC)$

Hence proved.

Tutorialspoint

e

Updated on: 10-Oct-2022

31 Views

Kickstart Your Career

Get certified by completing the course

Get Started