Question

a line through vertex c of ∆ABC bisects the median drawn from A prove that it divides the side AB in the ratio 1:2 ,AF/FB=1/2

any one solve it with full proff then I mark it as branlist answer.

Answer 1

Consider the attached diagram,

Consider G a point on BF such that DG ║CF
 that makes ΔBGD ~ΔBFC

Applying the following, 
In similar triangles, The line joining the points of two adjacent sides of a triangle is parallel to the third side. If two triangles are equiangular, their corresponding sides are proportional.

so BG/BF = BD/BC = 1/2
∵ BG = BF - GF
∴ (BF - GF)/BF = 1/2  or GF = BF/2

∵ DG║EF ∴ ΔAFE ~ΔAGD

Applying the following again,
In similar triangles, The line joining the points of two adjacent sides of a triangle is parallel to the third side. If two triangles are equiangular, their corresponding sides are proportional.

so AF/GF = AE/AD = 1/2
GF = 2AF
so AF/(FB) = 1/2