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Question
ABCD is a rhombus. EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced, meet at right angles.
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Solution
We know that the diagonals of a rhombus are perpendicular bisectors of each other
∴ OA = OC
OB = OD
∠AOD = ∠COD = 90°
And ∠AOB = ∠COB = 90°
In ΔBDE, A and O are midpoints of BE and BD, respectively
OA || DE
OC || DG
In ΔCFA, B and O are midpoints of AF and AC, respectively
∴ OB || CF
OD || GC
Thus, in quadrilateral DOCG, we have
OC || DG and OD || GC
⇒ DOCG is a parallelogram
∠DGC = ∠DOC
∠DGC = 90°
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