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ABCD is a rhombus. AB is produced to F and BA is produced to E such that AB = AE = BF. Then :
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- ED > CF
- ED ⊥ CF
- ED² + CF² = EF²
- ED || CF
- ED > CF
Correct Option: B
According to question , we draw a figure of rhombus ABCD 
We know that diagonals of a rhombus are perpendicular bisector of each other.
∴ OA = OC; OB = OD
∠AOD = ∠COD = 90°
∠AOB = ∠COB = 90°
In ∆ BDE,
OA || DE
⇒ OC || DG
In ∆CFA,
OB || CF
⇒ OD || GC
In quadrilateral DOCG ,
OC || DG
⇒ OD || GC
∴ DOCG is a parallelogram.
∴ ∠DGC = ∠DOC
⇒ ∠DGC = 90°
EG ⊥ GF or ED ⊥ CF
