# In figure below, $D$ and $\mathrm{E}$ are two points on $\mathrm{BC}$ such that $\mathrm{BD}=\mathrm{DE}=\mathrm{EC}$. Show that $\operatorname{ar}(\mathrm{ABD})=\operatorname{ar}(\mathrm{ADE})=\operatorname{ar}(\mathrm{AEC})$."

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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.

Updated on 10-Oct-2022 13:46:28