In a $Δ\ ABC$, $D$ and $E$ are points on $AB$ and $AC$ respectively, such that $DE\ ∥\ BC$. If $AD\ =\ 2.4\ cm$, $AE\ =\ 3.2\ cm$, $DE\ =\ 2\ cm$ and $BC\ =\ 5\ cm$. Find $BD$ and $CE$. "
Given:
In a $Δ\ ABC$, $D$ and $E$ are points on $AB$ and $AC$ respectively, such that $DE\ ∥\ BC$.
$AD\ =\ 2.4\ cm$, $AE\ =\ 3.2\ cm$, $DE\ =\ 2\ cm$ and $BC\ =\ 5\ cm$.
To do:
Here, we have to find $BD$ and $CE$.
Solution:
$DE || BC$, $AB$ is transversal.
$\angle APQ = \angle ABC$ (corresponding angles)
$DE || BC$, $AC$ is transversal.
$\angle AED = \angle ACB$ (corresponding angles)
In $\vartriangle ADE$ and $\vartriangle ABC$,
$\angle ADE=\angle ABC$
$\angle AED=\angle ACB$
$\vartriangle ADE = \vartriangle ABC$ (By AA similarity criteria)
$\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$ (CPCT)
$\frac{AD}{AB} = \frac{DE}{BC}$
$\frac{2.4} {(2.4 + DB)} = \frac{2}{5}$ ($AB = AD + DB$)
$2.4 + DB = \frac{2.4\times5}{2}$
$2.4 + DB = 6$
$DB = 6 – 2.4$
$DB = 3.6 cm$
Similarly,
$\frac{AE}{AC} = \frac{DE}{BC}$
$\frac{3.2}{(3.2 + EC)} = \frac{2}{5}$ ($AC = AE + EC$)
$3.2 + EC = \frac{3.2\times5}{2}$
$3.2 + EC = 8$
$EC = 8 – 3.2$
$EC = 4.8 cm$
Therefore,
$BD = 3.6 cm$ and $CE = 4.8 cm$.
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