D and E are respectively the points on the sides AB and AC of a $Δ A B C$ such that AB = 5.6 cm. AD = 1.4 cm. AC = 7.2 cm and AE = 1.8 cm. Show that $D E ∥ B C$ . [1 MARK]

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Solution

Solution: 1 Mark

We have,

AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm.

Now, $\frac{A D}{D B} = \frac{1.4}{5.6} = \frac{1}{4}$

$\frac{A E}{E C} = \frac{1.8}{7.2} = \frac{1}{4}$

$\Rightarrow \frac{A D}{D B} = \frac{A E}{E C}$

Thus, DE divides sides AB and AC of $Δ A B C$ in the same ratio.

$∴$ By the converse of Basic Proportionality Theorem, we have

$D E ∥ B C$

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D and E are respectively the points on the sides AB and AC of a $Δ A B C$ such that AB = 5.6 cm. AD = 1.4 cm. AC = 7.2 cm and AE = 1.8 cm. Then, $D E ∥ B C$ .

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D and E are respectively the points on the sides AB and AC of a $Δ A B C$ such that AB = 5.6 cm. AD = 1.4 cm. AC = 7.2 cm and AE = 1.8 cm. Show that $D E ∥ B C$ . [1 MARK]