Mathematics
The alongside figure shows a parallelogram ABCD in which AE = EF = FC. Prove that :
(i) DE is parallel to FB
(ii) DE = FB
(iii) DEBF is a parallelogram.
Rectilinear Figures
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Answer
Join BD. Let BD intersect AC at point O.
(i) We know that,
Diagonals of parallelogram bisect each other.
∴ OB = OD and OA = OC.
From figure,
⇒ OA = OC
⇒ OA - AE = OC - FC (As, AE = FC)
⇒ OE = OF.
In quadrilateral DEBF,
⇒ OB = OD and OE = OF.
Since, diagonals of quadrilateral DEBF bisect each other,
∴ DEBF is a parallelogram.
∴ DE || FB (Opposite sides of parallelogram are parallel.)
Hence, proved that DE || FB.
(ii) We know that,
Opposite sides of parallelogram are equal.
In parallelogram DEBF,
∴ DE = FB.
Hence, proved that DE = FB.
(iii) Since, one pair of opposite sides of quadrilateral DEBF is equal and parallel,
i.e. DE || FB and DE = FB,
∴ DEBF is a parallelogram.
Hence, proved that DEBF is a parallelogram.
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