# ABCD IS A PARALLELOGRAM, G IS THE POINT ON AB SUCH THAT AG = 2GB, E IS A POINT OF CD SUCH THAT CE = 2DE AND AND F IS A POINT OF BC SUCH THAT BF = 2FC. PROVE THAT1) ar(ADEG) = ar(GBCE), 2) ar(EGB) = 1/6 ar(ABCD) , 3) ar(EFC) = 1/2 ar(EBF) , 4) ar(EBG) = ar(EFC) , 5) FIND WHAT PORTION OF THE ARE OF PARALLELOGRAM IS THE AREA OF EFG.

1) Given: ABCD is a parallelogram in which AG=2GB,CE=2DE and BF=2FC.

To prove: ar( gmADEG) = ar( gmGBCE)

Proof: Since ABCD is a parallelogram, we have AB CD and AB =CD.

∴ BG= AB

and DE= CD= AB

∴   BG = DE

∴ ADEH is a gm    [since AH is parallel to DE and AD is parallel to HE]

ar( gmADEH )=ar( gmBCIG )...........(i)

[since DE=BG and AD=BC parallelogram with corresponding sides equal]

area ( ΔHEG)=area( ΔEGI).............(ii)

[diagonals of a parallelogram divide it into two equal areas]

From (i) and (ii), we get,

ar( gmADEH )+area ( ΔHEG)=ar( gmBCIG )+area( ΔEGI)

ar( gmADEG) = ar( gmGBCE) 2)

Height,  h of gm ABCD and ΔEGB is the same.

Base of ΔEGB = area(ABCD) = h AB
area(EGB) = area(EGB) = 3)  Let the distance between EH and CB = x ⇒ Area(EFC) =  Area(EBF)

4) If g = altitude from AD to BC
area(EFC) = ⇒ area(EFC) = area(EBG)

5) area(EFG) = area(EGB)+area(EBF)+area(EFC) • 11

AB=DC , AG=EC , GE(COMMON) , BC=AD IF ALL THE SIDES OF BOTH THE QUADRILATERAL ARE EQUAL TO EACH OTHER SO THE AREA OF BOTH QUADRILATERAL ARE EQUAL

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