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 th point of BC such that BF = 2FC. prove that : 1 ) ar( triangle EBG )= ar( triangle EFC) 2 ) find what portion of the area of parallelogram is the area of triangle EFG.
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Area (ΔEGB) = (1/2) Area (||gm ABCD)
AG = 2GB ( Given)
CE = 2DE ( Given)
BF = 2FC. ( Given)
AB =CD and AB || CD
BG = (1/3)AB and
DE= (1/3) CD= AB
Since, parallelogram ABCD and ΔEGB have same height
Let the height be = ‘h’
Base of ΔEGB = (1/3) AB
ar(||gm ABCD) = AB × h
Area of Δ EGB = 1/2 × bg × h
= 1/2 × 1/3 × AB × h
Therefore, area (ΔEGB) = (1/2) area (||gm ABCD)
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