ABCD is a parallelogram. G is a point on AB such that AG = 2GB and E is point on 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 ( )=1 6 ar(ABCD) iii. ar( )=1 2 ar( ) iv. ar( )=ar( ) v. Find what portion of the parallelogram is the area of .

ABCD is a parallelogram AG = 2GB, CE = 2DE, BF = 2FC To prove: ABCD is ||gm ar(EBG) = ar(EFC) AB | CD (AB = CD) BG=1 3 AB, DE=1 3 CD BG=DE ADEH is |gm ar( |gm ADEH) = ar( |gm BCIG) ...(i) ar( ) = ar( ) ..(ii) ( diagonal of |gm divided into 2 equal areas) (i) and (ii) ( |gm ADEG)=ar( |gm GBCE) Height, h of |gm ABCD and is Its same Base of =1 3 AB area of ABCD = h AB ar( )=1 6 1 3 AB=1 6 h ar(EGB)=1 6 Area(ABCD) let dislaver between EH and CB = x ar(EBF)=1 2 =1 2 2 3 BC =1 3 ar(EFC)=1 2 =1 2 1 3 BC =1 2 (EBF) (EFC)=1 2 area of EBF If g = altitute from AD to BC ar(EFC)=1 2 2 3 g 1 3 BC=1 9 ar(ABCD) (EFC)=2 3 ar EBG ar(EFG)=ar(EGB)+ar(FBF)+ar(EFC) =1 6 ABCD+2ar(EFC)+ar(EFC) =1 6 ABCD+3ar(EFC) =1 6 ABCD+ 3 1 9 = (1 6 +1 3 )ABCD = (1+2 6 )ABCD =1 2 ABCD

ABCD is a parallelogram. G is a point on AB such that AG = 2GB an…

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Question

$\text{ABCD}$ is a parallelogram. $G$ is a point on $AB$ such that $AG = 2GB$ and $E$ is point on $DC$ such that $CE = 2DE$ and $F$ is the point of $BC$ such that $BF = 2FC$. Prove that:
$i. \text{ar}(\text{ADEG})=\text{ar}(\text{GBCE})$
$ii. \text{ar}{(\triangle\text{EGB}})=\frac{1}{6}\text{ar}(\text{ABCD})$
$iii. \text{ar}(\triangle\text{EFC})=\frac{1}{2}\text{ar}(\triangle\text{EBF})$
$iv. \text{ar}(\triangle\text{EBG})=\text{ar}(\triangle\text{EFC})$
$v.$ Find what portion of the parallelogram is the area of $\triangle\text{EFG}.$

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Answer

$\text{ABCD}$ is a parallelogram $AG = 2GB, CE = 2DE, BF = 2FC$
To prove: $\text{ABCD}$ is $||gm$ ar$(EBG)$ = ar$(EFC)$
$AB \| CD$
$(AB = CD)$
$\text{BG}=\frac{1}{3}\text{AB},\ \text{DE}=\frac{1}{3}\text{CD}$
$\therefore\ \text{BG}=\text{DE}$
$\therefore\ \text{ADEH}$ is $\|gm$ ar$(\|gm \text{ADEH}) = ar(\|gm \text{BCIG}) ...(i)$
$ar(\triangle\text{HEG}) = ar(\triangle\text{EGI}) ..(ii)$
$(\because$ diagonal of $\|gm$ divided into $2$ equal areas$) (i)$ and $(ii)$
$\therefore\text{ar}(\|\text{gm}\text{ ADEG})=\text{ar}(\|\text{gm}\text{ GBCE})$
Height, $h$ of $\|gm$
$\text{ABCD}$ and $\triangle\text{EGB}$ is Its same Base of $\triangle\text{EGB}=\frac{1}{3}\text{AB}$
area of $\text{ABCD} = h$
$\times$
$AB$
$\text{ar}(\triangle\text{EGB})=\frac{1}{6}\times\text{h}\times\frac{1}{3}\text{AB}=\frac{1}{6}\text{h}\times\text{AB}$
$\text{ar}(\text{EGB})=\frac{1}{6}$Area$(\text{ABCD})$ let dislaver between $EH$ and $CB = x$
$\text{ar}(\text{EBF})=\frac{1}{2}\times\text{BF}\times\text{x}=\frac{1}{2}\times\frac{2}{3}\text{BC}\times\text{x}=\frac{1}{3}\times\text{BC}\times\text{x}$
$\text{ar}(\text{EFC})=\frac{1}{2}\times\text{CF}\times\text{x}=\frac{1}{2}\times\frac{1}{3}\text{BC}\times\text{x}=\frac{1}{2}\times\text{ar}\times(\text{EBF})$
$\Rightarrow\text{ar}(\text{EFC})=\frac{1}{2}\times$ area of $\text{EBF}$
If $g =$ altitute from $AD$ to $BC$
$\text{ar}(\text{EFC})=\frac{1}{2}\times\frac{2}{3}\text{g}\times\frac{1}{3}\text{BC}=\frac{1}{9}\text{ar}(\text{ABCD})$
$\Rightarrow\text{ar}(\text{EFC})=\frac{2}{3}\text{ar}\text{ EBG}$
$\text{ar}(\text{EFG})=\text{ar}(\text{EGB})+\text{ar}(\text{FBF})+\text{ar}(\text{EFC})$
$=\frac{1}{6}\text{ABCD}+2\text{ar}(\text{EFC})+\text{ar}(\text{EFC})$
$=\frac{1}{6}\text{ABCD}+3\text{ar}(\text{EFC})$
$=\frac{1}{6}\text{ABCD}+\not3\times\frac{1}{\not9}\times\text{ABCD}$
$=\Big(\frac{1}{6}+\frac{1}{3}\Big)\text{ABCD}$
$=\Big(\frac{1+2}{6}\Big)\text{ABCD}$
$=\frac{1}{2}\text{ABCD}$