Abcd is a parallelogram, g is the point on ab such that ag=2gb, e is a point on dc such that ce=2de and f is the point on bc such that bf=2fc. Prove that i) ar(ebg)=ar(efc)
Answers
) 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) = hAB
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)