ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC such that CE = 2 DE and F is the point of BC such that BF = 2 FC. Prove that: (i)ar(ADEG)=ar(GBCE)(ii)ar(△EGB)−16ar(ABCD)(iii)ar(△EFC)=12ar(△EBF)(iv)ar(△EGB)=32ar(△EFC) (v) Find what portion of the ara of parallelogram is the area of △EFG.

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

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 = 2 DE and F is the point of BC such that BF = 2 FC. Prove that: (i)ar(ADEG)=ar(GBCE)(ii)ar(△EGB)−16ar(ABCD)(iii)ar(△EFC)=12ar(△EBF)(iv)ar(△EGB)=32ar(△EFC) (v) Find what portion of the ara of parallelogram is the area of △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 = 2 DE and F is the point of BC such that BF = 2 FC. Prove that: $(i) a r (A D E G) = a r (G B C E) (i i) a r (△ E G B) - \frac{1}{6} a r (A B C D) (i i i) a r (△ E F C) = \frac{1}{2} a r (△ E B F) (i v) a r (△ E G B) = \frac{3}{2} a r (△ E F C)$ (v) Find what portion of the ara of parallelogram is the area of $△ E F G .$