In figure below, $DE\ ∥\ BC$ such that $AE\ =\ (\frac{1}{4})AC$. If $AB\ =\ 6\ cm$, find $AD$.
"
Given:
In the given figure, $DE\ ∥\ BC$ such that $AE\ =\ (\frac{1}{4})AC$ and $AB\ =\ 6\ cm$.
To do:
We have to find $AD$.
Solution:
In $\vartriangle ADE$ and $\vartriangle ABC$,
$\angle A = \angle A$ (Common)
$\angle ADE = \angle ABC$ ($AB||QR$, Corresponding angles)
Therefore,
$\vartriangle ADE ∼ \vartriangle ABC$ (By AA similarity)
$\frac{AD}{AB} = \frac{AE}{AC}$ (Corresponding parts of similar triangles are proportional)
$\frac{AD}{6} = \frac{1}{4}$ ($AE\ =\ (\frac{1}{4})AC$, this implies, $\frac{AE}{AC} =\frac{1}{4}$)
$AD = \frac{6}{4}$
$AD = 1.5\ cm$
The measure of $AD$ is $1.5\ cm$.
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