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Question
In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that: EGFH is a parallelogram.
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Solution
In quadrilateral AECF,
AE = CF ....[from (i)]
AE || CF ....[as AB || DC]
⇒ AECF is a parallelogram.
⇒ EC || AF or EH || GF ....(i)
In quadrilateral BFDE,
BE = DF ....[from (i)]
BE || DF ....[as AB || DC]
⇒ BEDF is a parallelogram
⇒ BF || ED or HF || EG ....(ii)
From (i) and (ii), we get that
EGFH is parallelogram.
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