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Question
In AABC, D and E are two points on the side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet the side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meet the side BC at points M and N respectively. Prove that BM = MN = NC.
Solution
In ΔAEG,
D is the mid-point of AE and DF || EG || BC
Therefore, F is the mid-point of AG.
⇒ AF = FG ....(1)
Again, DF || EG || BC.
Therefore, FG = GC ....(2)
Similarly, since GN || FM || AB,
therefore MB = MN = NC. ...(proved)
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