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.

Answer 1

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.