Question

1. In the figure, seg DE || side AC and
seg EF || side BA then prove that
AD/ AF*DB /FC=1​

Answer 1

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