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)

Question

Philoid · Accepted Answer

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

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 (

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)