Question

ABCD IS A PARALLELOGRAM, G IS THE POINT ON AB SUCH THAT AG = 2GB, E IS A POINT OF CD SUCH THAT CE = 2DE AND AND F IS A POINT OF BC SUCH THAT BF = 2FC. PROVE THAT
1) ar(EGB) = 1/6 ar(ABCD)
2)ar(EFC) = 1/2 ar(EBF)
3) ar(EBG) = ar(EFC)
4) FIND WHAT PORTION OF THE ARE OF PARALLELOGRAM IS THE AREA OF EFG.

Answer 1

1)
height h of ABCD and EGB is the same
base of EGB = 1/3 AB
area(ABCD) = h*AB
area(EGB) = 1/2 h*(1/3 AB) = 1/6 h*AB

2)
base of EFC = 1/2 base of EBF
so area is 1/2 as well

3)
if g = altitude from AD to BC
area(EFC) = 1/2 * (2/3 g) * (1/3 BC) = 1/9 area(ABCD)
So, I get area(EFC) = 2/3 area(EBG)
Is there a typo here?

4)
area(EFG) = area(EGB)+area(EBF)+area(EFC)
= 1/6(ABCD)+2area(EFC)+area(EFC)
= 1/6(ABCD) + 3(1/9 ABCD)
= 1/2(ABCD)