D and E are points on the sides AB and AC respectively of a $\triangle\text{ABC}$ such that DE || BC: If $\frac{\text{AD}}{\text{AB}}=\frac{8}{15}$ and EC = 3.5cm, find AE.
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In $\triangle\text{ABC},$ it is given that DE || BC.
Applying Thales' theorem, we get:
$\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}$
$\Rightarrow\frac{\text{8}}{\text{15}}=\frac{\text{AE}}{\text{AE}+\text{EC}}$
$\Rightarrow\frac{\text{8}}{\text{15}}=\frac{\text{AE}}{\text{AE}+\text{3.5}}$
$\Rightarrow8\text{AE}+28=15\text{AE}$
$\Rightarrow7\text{AE}=28$
$\Rightarrow\text{AE}=4\text{cm}$
Applying Thales' theorem, we get:
$\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}$
$\Rightarrow\frac{\text{8}}{\text{15}}=\frac{\text{AE}}{\text{AE}+\text{EC}}$
$\Rightarrow\frac{\text{8}}{\text{15}}=\frac{\text{AE}}{\text{AE}+\text{3.5}}$
$\Rightarrow8\text{AE}+28=15\text{AE}$
$\Rightarrow7\text{AE}=28$
$\Rightarrow\text{AE}=4\text{cm}$
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