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Question
In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively.
Prove that:
(i) Triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.
Solution
The figure is shown below
(i) From ΔHEB and ΔFHC
BE = FC
∠EHB = ∠FHC ...[ Opposite angle ]
∠HBE = ∠HFC
∴ ΔHEB ≅ ΔFHC
∴ EH = CH , BH = FH
(ii) Similarly AG = GF and EG = DG …..(1)
For triangle ECD,
F and H are the mid-point of CD and EC.
Therefore HF || DE and
HF = `[1]/[2]` DE ....(2)
From (1) and (2) we get,
HF = EG and HF || EG
Similarly, we can show that EH = GF and EH || GF
Therefore GEHF is a parallelogram.
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