Question
In a $\triangle\text{ABC}$ D and E are points on AB and AC respectively such that DE || BC. If AD = 2.4cm, AE = 3.2cm, DE = 2cm and BC = 5cm, find BD and CE.
✓
Answer
In the $\triangle\text{ABC},$ DE || BC
AD = 2.4cm, AE = 3.2cm, DE = 2cm and BC = 5cm
$\text{In}\ \triangle\text{ABC}$
$\because\text{DE}||\text{BC}$
$\therefore\triangle\text{ADE}\sim\triangle\text{ABC}$
$\Rightarrow\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}=\frac{\text{DE}}{\text{BC}}$
$\Rightarrow\frac{2.4}{\text{AB}}=\frac{3.2}{\text{AC}}=\frac{2}{5}$
Now, $\frac{2.4}{\text{AB}}=\frac{2}{5}\Rightarrow\text{AB}=\frac{2.4\times5}{2}=6\text{cm}$
$\text{DB}=\text{AB}-\text{AD}=6.0-2.4=3.6\text{cm}$
And $\because\frac{3.2}{\text{AC}}=\frac{2}{5}\Rightarrow\text{AC}=\frac{3.5\times5}{2}=8.75\text{cm}$
$\therefore\text{CE}=\text{AC}-\text{AE}=8.0-3.2=4.8\text{cm}=8\text{cm}$
AD = 2.4cm, AE = 3.2cm, DE = 2cm and BC = 5cm
$\text{In}\ \triangle\text{ABC}$
$\because\text{DE}||\text{BC}$
$\therefore\triangle\text{ADE}\sim\triangle\text{ABC}$
$\Rightarrow\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}=\frac{\text{DE}}{\text{BC}}$
$\Rightarrow\frac{2.4}{\text{AB}}=\frac{3.2}{\text{AC}}=\frac{2}{5}$
Now, $\frac{2.4}{\text{AB}}=\frac{2}{5}\Rightarrow\text{AB}=\frac{2.4\times5}{2}=6\text{cm}$
$\text{DB}=\text{AB}-\text{AD}=6.0-2.4=3.6\text{cm}$
And $\because\frac{3.2}{\text{AC}}=\frac{2}{5}\Rightarrow\text{AC}=\frac{3.5\times5}{2}=8.75\text{cm}$
$\therefore\text{CE}=\text{AC}-\text{AE}=8.0-3.2=4.8\text{cm}=8\text{cm}$
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