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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: ΔHEB ≅ ΔHFC
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Solution
Since ABCD is a parallelogram,
AB = CD and AD = BC
Now, E and F are the mid-points of AB and CD respectively,
⇒ AE = EB = DF = FC ....(i)
In ΔHEB and ΔHFC,
BE = FC ....[From (i)]
∠EHB = ∠FHC ....(vertically opposite angles)
∠HBE = ∠HFC ....(Alternate interior angles)
∴ ΔHEB ≅ ΔHFC.
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