`A B C D` is a rhombus, `E A B F` is a straight line such that `E A=A B=B F` . Prove that `E D` and `F C` when produced meet at right angles.

Explanation

We know that the diagonals of a rhombus are perpendicular bisector of each other.
`OA=OC,OB=OD`
`/_AOD=/_COD=90^0`
`/_AOB=/_COB=90^0`
In `/_\BDE`, A and O are mid - points of BE and
BD respectively.
`OA∥DE`
`=>OC∥DG`
In `/_\CFA`,B and O are mid - points AF and AC respectively
`OB∥CF`
`OD∥GC`
Thus, in quadrilateral` DOCG`
`OC∥DG \and\ OD∥GC`
`=>DOCG` is a parallelogram.
`/_DGC=/_DOC`
`/_DGC=90^0`