In the figure, $ABC$ and $BDC$ are two equilateral triangles such that $D$ is the mid-point of $BC$. $AE$ intersects $BC$ in $F$.
Prove that $ \operatorname{ar}(\Delta \mathrm{BDE})=\frac{1}{2} \operatorname{ar}(\Delta \mathrm{BAE}) $.
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Given:

$ABC$ and $BDC$ are two equilateral triangles such that $D$ is the mid-point of $BC$. $AE$ intersects $BC$ in $F$.

To do:

We have to prove that \( \operatorname{ar}(\Delta \mathrm{BDE})=\frac{1}{2} \operatorname{ar}(\Delta \mathrm{BAE}) \).

Solution:

Let $\mathrm{AB}=\mathrm{BC}=\mathrm{CA}=x$

This implies,

$\mathrm{BD}=\frac{x}{2}$             ($\mathrm{D}$ is the mid point of $\mathrm{BC}$)

$\triangle \mathrm{ABC}$ and $\triangle \mathrm{BED}$ are equilateral triangles.

$\Rightarrow \angle \mathrm{ACB}=\angle \mathrm{DBE}=60^{\circ}$

$\angle \mathrm{ACB}$ and $\angle \mathrm{DBE}$ are alternate angles.

Therefore,

$\mathrm{AB} \| \mathrm{DE}$

$ar(\Delta \mathrm{BAE})=2 a r(\Delta \mathrm{BEC})$

$=2 \operatorname{ar}(\Delta \mathrm{BDE})$          ($\mathrm{ED}$ is a median of $\Delta \mathrm{EBC}$)

$\Rightarrow \operatorname{ar}(\Delta \mathrm{BDE})=\frac{1}{2} \operatorname{ar}(\Delta \mathrm{BAE})$

Hence proved. 

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Updated on: 10-Oct-2022

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