ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, 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 (ADEG) = ar (GBCE)(ii) ar (Δ EGB) = 16 ar (ABCD)(iii) ar (Δ EFC) = 12 ar (Δ EBF)(iv) ar (Δ EBG) = ar (Δ EFC)(v) Find what portion of the area of parallelogram is the area of Δ EFG

ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E

1.	ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, 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 (ADEG) = ar (GBCE)(ii) ar (Δ EGB) = 16 ar (ABCD)(iii) ar (Δ EFC) = 12 ar (Δ EBF)(iv) ar (Δ EBG) = ar (Δ EFC)(v) Find what portion of the area of parallelogram is the area of Δ EFG
Answer» ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, 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 (ADEG) = ar (GBCE) (ii) ar (Δ EGB) = $\frac{1}{6}$ ar (ABCD) (iii) ar (Δ EFC) = $\frac{1}{2}$ ar (Δ EBF) (iv) ar (Δ EBG) = ar (Δ EFC) (v) Find what portion of the area of parallelogram is the area of Δ EFG