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प्रश्न
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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उत्तर

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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संबंधित प्रश्न
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see the given figure). Show that F is the mid-point of BC.

Use the following figure to find:
(i) BC, if AB = 7.2 cm.
(ii) GE, if FE = 4 cm.
(iii) AE, if BD = 4.1 cm
(iv) DF, if CG = 11 cm.

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