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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

(i) BDEF is a parallelogram.

(ii) ar (DEF) = ar (ABC)

(iii) ar (BDEF) = ar (ABC)

Answer :

Kindly refer the following link :

https://www.meritnation.com/ask-answer/question/d-e-and-f-are-respectively-the-mid-points-of-the-sides-bc/areas-of-parallelograms-and-triangles/6294661

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THE QUESTION IS NOT CLEAR.

1ST PART IS CLEAR

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proof - in BFED

DE//BA

DF=1/2BA

SO,DE//BF AND DE=BF [OPP.SIDES ARE EQUAL AND // IS A //gm

SIMILARLY AEDF AND CEED ARE //gm

BFED,AEDF,CEFD ARE //gm

ar(DEF)=ar(BFD) [DIAGONALS OF A //gm

DIVIDES IT INTO TWO TRIANGLES]

ar(DEF)=ar(AEF)

ar(DEF)=ar(CDE)

hence the proof.....

  • -7

after ar(DEF)=ar(CDE)

ar(def)+ar(aef)+

ar(bdf)+ar(cde)+ar(abc)

ar(def)+ar(def)+

ar(def)+ar(def) = ar(abc)

4ar(def) = ar(abc)

ar(def) = 1/4 ar(abc)

ar(bdef) = 2ar(def)

ar(bdef) = 2x1/4 ar(abc) [proved]

ar(bdef) = 1/2 ar(abc)

hence the proof

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