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 THAT
1) 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.
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)
Height, h of gm ABCD and ΔEGB is the same.
Base of ΔEGB =
area(ABCD) = hAB
3) Let the distance between EH and CB = x
⇒ Area(EFC) = Area(EBF)
4) If g = altitude from AD to BC
⇒ area(EFC) = area(EBG)
5) area(EFG) = area(EGB)+area(EBF)+area(EFC)