D- E and F are respectively the mid-points of the sides BC- CA and AB of a - A B C - Show thata BDEF is a parallelogram-b ar DEF - 1 - 4 ar ABC c ar BDEF - 1 - 2 ar ABC

To prove:(i) BDEF is parallelogram #160;(i) ar ( DEF)=14ar(Δ ABC) (ii) ar( BDEF)=12ar Delta ABC) Proof : #160;(i) Δ ABC , ∵ F is t ...

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Byju's Answer

Standard IX

Mathematics

The Mid-Point Theorem

D, E and ...

Question

$D$ , $E$ and $F$ are respectively the mid-points of the sides $B C$ , $C A$ and $A B$ of a $Δ A B C$ .
Show that
(a) $B D E F$ is a parallelogram.
(b) $ar (D E F) = \frac{1}{4} ar (A B C)$
(c) $ar (B D E F) = \frac{1}{2} ar (A B C)$

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Solution

To prove:
(i) $□ B D E F$ is parallelogram
(i) $a r (□ D E F) = \frac{1}{4} a r (Δ A B C)$
(ii) $a r (□ B D E F) = \frac{1}{2} a r D e l t a A B C)$

Proof :
(i) $Δ A B C$ ,

$∵ F$ is the mid point of side $A B$ and $E$ is the mid point of side $A C$ .

$∴ E F | | B C$

$∵$ In a triangle, the line segment joining the mid points of any two sides is parallel to the third side.

$\Rightarrow E F | | B D$ ...(1)

Similarly, $E D | | B F$ ...(2)

In view of (1) and (2),

$□ B D E F$ is a parallelogram.

$∵$ A quadrilateral is a parallelogram if its opposite sides are parallel
(ii) As in (1), we can prove that

$□ A F D E$ and $□ F D C E$ are parallelogram.

$∴ F D$ is a diagonal of $| |^{g m}$ $B D E F$

$a r (Δ F B D) = a r (Δ D E F)$ ....(3)

Similarly, $a r (Δ D E F) = a r (Δ F A E)$ ...(4) and,

$a r (Δ D E F) = a r (Δ D E E)$ ...(5)

From (3), (4) and (5), we have $a r (Δ F B D) = a r (Δ D E F)$

$= a r (Δ F A E) = a r (Δ D E C)$ ...(6)

$∵ Δ A B C$ is divided into four non-overlapping triangles

$Δ F B D, Δ D E F, Δ F A E$ and $Δ D C E$ .

$∴ a r (Δ A B C) = a r (Δ F B D) + a r (Δ D E F) + a r (Δ F A E) + a r (Δ D C E) = 4 a r (Δ D E F)$ | .....from (6)

$\Rightarrow a r (Δ D E F) = 1 / 2 a r c (Δ A B C)$ ...(7)

(iii) $a r (□ B D E F)$

$= a r (Δ F B D) + a r (Δ D E F)$

$= a r (Δ D E F) + a r (Δ D E F)$

from (3)
$= 2 a r (Δ D E F)$

$= 2. \frac{1}{4} a r (Δ A B C)$ .......from (7)

$= \frac{1}{4} a r (Δ A B C)$

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Similar questions

D, E

and

F

are respectively the mid-points of the sides

B C, C A

and

A B

of a

△ A B C

. Show that
(i)

B D E F

is a parallelogram
(ii)

a r (D E F) = \frac{1}{4} a r (A B C)

(iii)

a r (B D E F) = \frac{1}{2} a r (A B C)

D, E and F are the mid point of sides BC , CA,AB of a triangle ABC
1. BDEF is a parallelogram
2. $a r (D E F) = \frac{1}{4} a r (A B C)$
3. $(B D E F) = \frac{1}{2} a r (A B C)$

Q. If

D, E, F

are the mid-points of the sides

B C, C A

and

A B

respectively of

△ A B C

, prove that

B D E F

is a parallelogram. And also show that

a r (△ D E F) = \frac{1}{4} a r (△ A B C)

D, E

and

F

are respectively the mid-point of the sides

B C, C A

and

A B

of a

△ A B C

Show that

a r (B D E F) = \frac{1}{2} a r (A B C)

Q. In the figure

△ A B C, D, E, F

are the midpoint of sides

B C, C A

and

A B

respectively. Show that

a r (B D E F) = \frac{1}{2} a r (△ A B C)

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D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC . Show that(a)BDEF is a parallelogram.(b)ar(DEF)=14ar(ABC)(c)ar(BDEF)=12ar(ABC)

$D$ , $E$ and $F$ are respectively the mid-points of the sides $B C$ , $C A$ and $A B$ of a $Δ A B C$ .
Show that
(a) $B D E F$ is a parallelogram.
(b) $ar (D E F) = \frac{1}{4} ar (A B C)$
(c) $ar (B D E F) = \frac{1}{2} ar (A B C)$