# ABC is a triangle. AD is the median and E is the mid point of AD.BE is joined an produced to intersect AC at F.Prove that AF is equal to one-third of AC

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by harsh10

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by harsh10

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so, bd=dc

consider tri.BCF

d is the m.p of side bc

draw a line DG from D parallel to BF

according to M.P.theorem, CG=FG..................(1)

now,consider tri.ADG

e and f are m.p's of sides AD and AG respectively

therefore AF=FG.............................................(2)

from 1 and 2 we get

AF=FG=CG

i.e,AF is 1/3 of the side AC