ABCD is a rhombus, EAFB is a straight line such that EA = AB = BF.
Question:

ABCD is a rhombus, EAFB is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.

Solution:

We know that the diagonals of a rhombus are perpendicular bisector of each other.

∴ OA = OC, OB = OD, and ∠AOD = ∠COD = 90°

And ∠AOB = ∠COB = 90°

In ΔBDE, A and O are mid-points of BE and BD respectively.

OA ∥ DE

OC ∥ DG

In ΔCFA, B and O are mid-points of AF and AC respectively.

OB ∥ CF

OD ∥ GC

Thus, in quadrilateral DOGC, we have

OC ∥ DG and OD ∥ GC

⟹ DOCG is a parallelogram

∠DGC = ∠DOC

∠DGC = 90°

 

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