Question
In a $\triangle\text{ABC, D}$ and $E$ are points on the sides $AB$ and $AC$ respectively. For the following cases show that $DE || BC:$
$AB = 5.6\ cm, AD = 1.4\ cm, AC = 7.2$ and $AE = 1.8\ cm$.

Answer

Given $AB = 5.6\ cm, AD = 1.4\ cm, AC = 7.2\ cm$ and $AE = 1.8\ cm$
Now, $\frac{\text{AD}}{\text{AB}}=\frac{1.4}{5.6}=\frac{1}{4}$
And, $\frac{\text{AE}}{\text{AC}}=\frac{1.8}{7.2}=\frac{1}{4}$
$\therefore\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}$
Thus, $DE$ divides sides $AB$ and $AC$ of $\triangle\text{ABC}$ in the same ratio. Therefore, by converse of basic proportionality theorem, we have $DE || BC.$

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