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Question
In triangle ABC, D and E are points on side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meets side BC at points M and N respectively. Prove that: BM = MN = NC.
Solution
The figure is shown below
AD = DE = EB
In ΔAEG
AD = DE & DF || EG
Using mid point theorem
F is midpoint of AG
∴ AF = FG ...(1)
DF || EG || BC and DE = BE,
∴ FG = GC ...(2)
(1), (2) we get
AF = GF = GC
Similarly since GN || FM || AB
∴ BM = MN = NC
Hence, proved.
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