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In figure below, $D$ and \( \mathrm{E} \) are two points on \( \mathrm{BC} \) such that \( \mathrm{BD}=\mathrm{DE}=\mathrm{EC} \). Show that \( \operatorname{ar}(\mathrm{ABD})=\operatorname{ar}(\mathrm{ADE})=\operatorname{ar}(\mathrm{AEC}) \).
Given:
$D$ and \( \mathrm{E} \) are two points on \( \mathrm{BC} \) such that \( \mathrm{BD}=\mathrm{DE}=\mathrm{EC} \).
To do:
We have to show that \( \operatorname{ar}(\mathrm{ABD})=\operatorname{ar}(\mathrm{ADE})=\operatorname{ar}(\mathrm{AEC}) \).
Solution:
In $\triangle ABE$
$BD=DE$
This implies,
$AD$ is the median.
We know that,
The median of a triangle divides it into two parts of equal areas.
This implies,
$ar(\triangle ABD) = ar(\triangle AED)$.........(i)
In $\triangle ADC$,
$DE=EC$
$AE$ is the median
This implies,
$ar(\triangle ADE) = ar(\triangle AEC)$..........(ii)
From (i) and (ii), we get,
$ar(\triangle ABD) = ar(\triangle ADE) = ar(\triangle AEC)$
Hence proved.