If $D$ is a point on the side $AB$ of $\triangle ABC$ such that $AD : DB = 3:2$ and $E$ is a point on $BC$ such that $DE\parallel AC$. Find the ratio of areas of $\triangle ABC$ and $\triangle BDE$.
Given:
$D$ is a point on the side $AB$ of $\triangle ABC$ such that $AD : DB = 3:2$ and $E$ is a point on $BC$ such that $DE\parallel AC$.
To do:
We have to find the ratio of areas of $\triangle ABC$ and $\triangle BDE$.
Solution:
$\frac{AD}{DB}=\frac{3}{2}$
Let $AD$ be $3x$ and $DB$ be $2x$. This implies,
$AB=AD+DB=3x+2x=5x$
In $\triangle BDE$ and $\triangle BAC$,
$\angle DBE=\angle ABC$ (Common angle)
$\angle BDE=\angle BAC$ (Corresponding angles)
Therefore,
$\triangle BDE \sim\ \triangle BAC$ (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,
$\frac{ar(\triangle ABC)}{ar(\triangle BDE)}=\frac{(AB)^2}{(DB)^2}$
$=\frac{(5x)^2}{(2x)^2}$
$=\frac{25x^2}{4x^2}$
$=\frac{25}{4}$
The ratio of areas of $\triangle ABC$ and $\triangle BDE$ is $25:4$.
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