1. In the figure, seg DE || side AC and
seg EF || side BA then prove that
AD/ AF*DB /FC=1
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Answer.
Given AD is the median of ΔABC and E is the midpoint of AD
Through D, draw DG || BF
In ΔADG, E is the midpoint of AD and EF || DG
By converse of midpoint theorem we have
F is midpoint of AG and AF = FG → (1)
Similarly, in ΔBCF
D is the midpoint of BC and DG || BF
G is midpoint of CF and FG = GC → (2)
From equations (1) and (2), we get
AF = FG = GC → (3)
From the figure we have, AF + FG + GC = AC
AF + AF + AF = AC [from (3)]
3 AF = AC
AF = (1/3) AC
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