Question:
In the given figure, AD is a median of ∆ABC and E is the mid-point of AD. If BE is joined and produced to meet AC in F, then AF = ?
(a) $\frac{1}{2} A C$
(b) $\frac{\overline{1}}{3} A C$
(3) $\frac{2}{3} A C$
(4) $\frac{3}{4} A C$
Solution:
(b) $1 / 3 A C$
Explanation:
Let G be the mid point of FC. Join DG.
In ∆BCF, D is the mid point of BC and G is the mid point of FC.
∴ DG || BF
⇒ DG || EF
In ∆ ADG, E is the mid point of AD and EF || DG.
i.e., F is the mid point of AG.
Now, AF = FG = GC [∵ G is the mid point of FC]
$\therefore A F=1 / 3 A C$
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