Question

In the adjoining figure, in triangle ABC, AD is the median through A and E is the mid-point of AD If BE produced meets AC in F, prove that AF = 1/3 AC.

Answer 1

Answer:

I hope this will clear yur doubt.

Answer 2

AnswEr:

Through D, draw DK || BF. In ∆ADK, E is the mid-point of AD and EK || DK.

$\\ \therefore \: \sf \: F \: is \: the \: mid - point \: of \: AK \\ \\ \implies \qquad \sf \: AF = FK \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1)$

In ∆BCF, D is the mid-point of BC and DK || BF.

$\therefore \sf \: \: K \: is \: the \: mid \: - \: point \: of \: FC \\ \\ \therefore \sf \: FK = KC \: \: \: \: \: \: \: \: \: \: \: \: \: \: (2)$

From (1) and (2), we have

$\qquad \sf \: AF = FK +KC \: \: \: \: \: \: \: \: \: \: \: \: \: \: (3) \\ \\ \\ \sf \: now \: \: . \: \: AC = AF + FK + KC \\ \\ \implies \sf \: AC =AF + AF + AF \qquad(using \: 3) \\ \\ \implies \sf \: AC = 3(AF) \\ \\ \qquad \sf \: AF = \frac{1}{3} \: AC$