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:
AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and EC = 5.5cm.
AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and EC = 5.5cm.
✓
Answer
We have,DE || BC
We have, AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and EC = 5.5cm
Now $\frac{\text{AD}}{\text{BD}}=\frac{5.7}{9.5}=\frac{57}{95}$
$\Rightarrow\frac{\text{AD}}{\text{BD}}=\frac{3}{5}$
And $\frac{\text{AE}}{\text{EC}}=\frac{3.3}{5.5}=\frac{33}{55}$
$\Rightarrow\frac{\text{AE}}{\text{EC}}=\frac{3}{5}$
Thus DE divides sides AB and AC of $\triangle\text{ABC}$ in the same ratio.Therefore, by the converse of basic proportionality theorem. We have DE || BC.
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