$D and E are points on sides AB and AC, such that AB = 5.6, AD = 1.4, AC = 7.2 and AE = 1.8 . Show that DE ∥ BC .$

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Solution

$G iven : AD = 1.4, AB = 5.6 ∴ DB = AB - AD = 5.6 - 1.4 = 4.2 AC = 7.2, AE = 1.8 ∴ EC = AC - AE = 7.2 - 1.8 = 5.4 In △ ABC, \frac{AE}{EC} = \frac{1.8}{5.4} = \frac{1}{3} \frac{AD}{DB} = \frac{1.4}{4.2} = \frac{1}{3} ∴ \frac{AE}{EC} = \frac{AD}{DB} = \frac{1}{3} Therefore, according to the converse of basic proportionality theorem, DE ∥ BC . Hence proved .$

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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]

D and E are respectively the points on the sides AB and AC of a $Δ$ ABC such that AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm, prove that DE || BC.

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 points on sides AB and AC, such that AB=5.6, AD=1.4,AC=7.2 and AE=1.8 .Show that DE∥BC.

Given: AD=1.4 , AB=5.6 ∴DB=AB-AD=5.6-1.4=4.2AC=7.2 , AE=1.8∴EC=AC-AE=7.2-1.8=5.4In △ABC,AEEC=1.85.4=13ADDB=1.44.2=13∴AEEC=ADDB=13Therefore, according to the converse of basic proportionality theorem, DE∥BC.Hence proved.

$D and E are points on sides AB and AC, such that AB = 5.6, AD = 1.4, AC = 7.2 and AE = 1.8 . Show that DE ∥ BC .$