If $\triangle ABC$ and $\triangle BDE$ are equilateral triangles, where $D$ is the mid-point of $BC$, find the ratio of areas of $\triangle ABC$ and $\triangle BDE$.
Given:
$\triangle ABC$ and $\triangle BDE$ are equilateral triangles, where $D$ is the mid-point of $BC$.
To do:
We have to find the ratio of areas of $\triangle ABC$ and $\triangle BDE$.
Solution:
In $\triangle ABC$ and $\triangle ABC$,
$\angle A=\angle E$ ($\triangle ABC$ and $\triangle BDE$ are equilateral triangles)
$\angle ABC=\angle BED$ ($\triangle ABC$ and $\triangle BDE$ are equilateral triangles)
Therefore,
$\triangle ABC \sim\ \triangle BDE$ (By AA similarity)
We know that,
If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.
Therefore,
$\frac{ar(\triangle ABC)}{ar(\triangle BDE)}=\frac{BC^2}{BD^2}$
$=\frac{(2BD)^2}{BD^2}$ ($D$ is the mid-point of $BC$)
$=\frac{4BD^2}{BD^2}$
$=\frac{4}{1}$
The ratio of areas of $\triangle ABC$ and $\triangle BDE$ is $4:1$.
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