Question

In ∆ABC, AD is the median through A and E is the midpoint of AD. BE produced meets AC in F. prove that AF = 1/3AC.

Answer 1

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Here's your answer ❗

Through D draw DG || BF.

In ∆ADG, E is the midpoint of AD and EF || DG.

∴ EF bisects AG, i.e., F is the mid - pt. of AG.

AF = FG

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

∴ DG bisects FC, i.e., G is the mid-point of FC.

i.e., FG = GC

from (i) and (ii) AF = FG = GC

Now, AC = AF + FG + GC

= AF + AF + AF = 3AF [from (iii)]

⇒ AF = 1/3 AC.

Hence, proved by vaibhav....

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Answer 2

The answer is in the pic