ABCD is a parallelogram, G is the point on AB such that AG = 2GB, E is a point of DC such that CE = 2DE and F is the point of BC such that BF=2FC. Prove that:

(i) ar(Δ ADGE) = ar(Δ GBCE)


(ii) ar(Δ EGB) = ar(Δ ABCD)


(iii) ar(Δ EFC) = ar(Δ EBF)


(iv) ar(Δ EBG) = ar(Δ EFC)

Given: ABCD is a parallelogram in which

AG = 2 GB


CE = 2 DE


BF = 2 FC


(i) Since ABCD is a parallelogram, we have AB ‖ CD and AB = CD


Therefore,


BG = AB


And,


DE = CD = AB


Therefore,


BG = DE


ADEH is a parallelogram (Since, AH is parallel to DE and AD is parallel to HE)


Area of parallelogram ADEH = Area of parallelogram BCIG (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,


Area of parallelogram ADEH + Area (ΔHEG) = Area of parallelogram BCIG + Area (ΔEGI)


Therefore,


Area of parallelogram ADEG = Area of parallelogram GBCE


(ii) Height, h of parallelogram ABCD and ΔEGB is the same


Base of ΔEGB = AB


Area of parallelogram ABCD = h * AB


Area (EGB) = * AB * h


= (h) * AB


= * Area of parallelogram ABCD


(iii) Let the distance between EH and CB = x


Area (EBF) = * BF * x


= * BC * x


= * BC * x


Area (EFC) = * CF * x


= * * BC * x


= * Area (EBF)


Area (EFC) = * Area (EBF)


(iv) As, it has been proved that


Area (EGB) = = * Area of parallelogram ABCD (iii)


Area ( = Area ()


Area ( = * * CE * EP


= * * * CD * EP


= * * Area of parallelogram ABCD


Area ( = * Area ( [By using (iii)]


Area ( = Area (


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