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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
Solution
We know that the diagonals of a rhombus are perpendicular bisector of each other
∴ OA = OC, OB = OD, `∠`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 DOCG, we have
OC || DG and OD || GC
⇒ DOCG is a parallelogram
`∠`DGC = `∠`DOC
`∠`DGC = 90°
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