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Standard X

Mathematics

Basic Proportionality Theorem

D and E are...

Question

$D$ and $E$ are respectively the points on the sides $A B$ and $A C$ of a $△ A B C$ such that $A B = 2.6 c m, A D = 1.3 c m$ and $A E = 1.5 c m$ ,Show that $D E ∥ B C$

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Solution

Here $\frac{A B}{A D} = \frac{2.6}{1.3} = 2$

$\frac{A C}{A E} = \frac{3}{1.5} = 2$

$∴ \frac{A B}{A D} = \frac{A C}{A E}$

By the converse of BPT,

$D E ∥ B C$

Hence proved.

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Similar questions

Q. In given figure

D

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E

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A B

and

A C

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and

A E = 1.8 c m

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Q. In

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Q. 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

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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D and E are respectively the points on the sides AB and AC of a △ABC such that AB=2.6 cm,AD=1.3 cm and AE=1.5 cm,Show that DE∥BC

Here ABAD=2.61.3=2ACAE=31.5=2∴ABAD=ACAEBy the converse of BPT,DE∥BCHence proved.

$D$ and $E$ are respectively the points on the sides $A B$ and $A C$ of a $△ A B C$ such that $A B = 2.6 c m, A D = 1.3 c m$ and $A E = 1.5 c m$ ,Show that $D E ∥ B C$

Here $\frac{A B}{A D} = \frac{2.6}{1.3} = 2$

$\frac{A C}{A E} = \frac{3}{1.5} = 2$

$∴ \frac{A B}{A D} = \frac{A C}{A E}$

By the converse of BPT,

$D E ∥ B C$

Hence proved.