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Standard X

Mathematics

Basic Proportionality Theorem

D and E are r...

Question

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.

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Solution

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

$∴$ BD = AB – AD = 5.6 – 1.4 = 4.2 cm

And EC = AC – AE = 7.2 – 1.8 = 5.4 cm

Now, AD / DB =1.4 / 4.2 = 1/3

And AE / EC = 1.8 / 5.4 = 1/3

$\Rightarrow$ AD / DB = AE / EC

DE || BC (by converse of basic proportionality theorem)

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Similar questions

Q. In given figure

D

and

E

are respectively the points on the sides

A B

and

A C

of a

△ A B C

such that

A B = 5.6 c m, A D = 1.4 c m, A C = 7.2 c m

and

A E = 1.8 c m

. Then show that

D E | | B C

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 $Δ A B C$ such that AB = 5.6 cm. AD = 1.4 cm. AC = 7.2 cm and AE = 1.8 cm. Then, $D E ∥ B C$ .

In a $Δ A B C$ , D and E are points on the sides AB and AC respectively. For each of the following cases show that DE || BC:
(i) AB = 12cm, AD = 8 cm, AE = 12 cm and AC= 18cm.
(ii)AB = 5.6cm, AD= 1.4cm, AC = 7.2 cm and AE= 1.8 cm.
(iii) AB= 10.8 cm, BD= 4.5cm, AC = 4.8cm and AE =2.8 cm.
(iv)AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and EC = 5.5 cm.

Q. In a

△

A B C, D

and

E

are points on the sides AB and AC respectively. For each of the following cases show that

D E

∥

B C

(i)

A B = 12 c m, A D = 8 c m, A E = 12 c m a n d A C = 18 c m

(ii)

A B = 5.6 c m, A D = 1.4 c m, A C = 7.2 c m a n d A E = 1.8 c m

(iii)

A B = 10.8 c m, B D = 4.5 c m, A C = 4.8 c m a n d A E = 2.8 c m

(iv)

A D = 5.7 c m, B D = 9.5 c m, A E = 3.3 c m a n d E C = 5.5 c m .

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

AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm ∴BD = AB – AD = 5.6 – 1.4 = 4.2 cm And EC = AC – AE = 7.2 – 1.8 = 5.4 cm Now, AD / DB =1.4 / 4.2 = 1/3 And AE / EC = 1.8 / 5.4 = 1/3 ⇒AD / DB = AE / EC DE || BC (by converse of basic proportionality theorem)

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.

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

$∴$ BD = AB – AD = 5.6 – 1.4 = 4.2 cm

And EC = AC – AE = 7.2 – 1.8 = 5.4 cm

Now, AD / DB =1.4 / 4.2 = 1/3

And AE / EC = 1.8 / 5.4 = 1/3

$\Rightarrow$ AD / DB = AE / EC

DE || BC (by converse of basic proportionality theorem)