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}(\triangle \mathrm{BDE})=\frac{1}{4} \operatorname{ar}(\Delta \mathrm{ABC}) $.
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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 \( \operatorname{ar}(\triangle \mathrm{BDE})=\frac{1}{4} \operatorname{ar}(\Delta \mathrm{ABC}) \).

Solution:

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

This implies,

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

Area of equilateral triangle $\mathrm{ABC}=\frac{\sqrt{3}}{4}(\text { side })^{2}$

$=\frac{\sqrt{3}}{4} x^{2} \mathrm{~cm}^{2}$

Area of equilateral triangle $\mathrm{BED}=\frac{\sqrt{3}}{4}(\frac{x}{2})^{2}$

$=\frac{\sqrt{3}}{4} \times \frac{x^{2}}{4}$

$=\frac{1}{4}(\frac{\sqrt{3}}{4} x^{2})$

$=\frac{1}{4}(\text { area of } \triangle \mathrm{ABC})$

Hence proved.

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Updated on: 10-Oct-2022

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