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Asked by  | 16th Feb, 2009, 03:28: PM

Expert Answer:

 

EA = AB = BF   (Given)

Now AB = BC = CD = DA    (sides of a rhombus)

BC = BF    

BFC = BCF   (angle opposite to equal side)

Similarly,

AED = ADE  (angle opposite to equal side)

ABC = BFC +BCF            (exterior angle)

             = 2BFC

DAB= ADE +AED             (exterior angle)

             = 2AED

2AED + 2BFC= 1800                (composite pairs)

AED + BFC= 900 

Now, In AFG

AED + BFC + EGF= 1800

EGF= 900

 

 

 

Answered by  | 17th Feb, 2009, 11:38: AM

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