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Standard X

Mathematics

Converse of Basic Proportionality Theorem

In given figu...

Question

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 $D E | | B C$ .

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Solution

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

$∴$ $B D = A B - A D = (5.6 - 1.4)$ cm $=$ $4.2$ cm
and ,

$E C = A C - A E = (7.2 - 1.8)$ cm $=$ $5.4$ cm

Now, $\frac{A D}{D B} = \frac{1.4}{4.2} = \frac{1}{3}$ and $\frac{A E}{E C} = \frac{1.8}{5.4} = \frac{1}{3}$

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

Thus, DE divides sides AB and AC of $△ A B C$ in the same ratio.
Therefore, by the converse of Basic Proportionality Theorem, we have $D E | | B C$

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Similar questions

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.

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 .

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

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

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.

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In given figure D and E are respectively the points on the sides AB and AC of a △ABC such that AB=5.6 cm,AD=1.4 cm,AC=7.2 cm and AE=1.8 cm. Then show that DE||BC.

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 $D E | | B C$ .