prove that AF=1/3 AC

Q. Prove that: AF=1/3 AC

draw a line DG parallel to BF.
In A D G,
E F parallel to DG
E is the mid point of AD
so, F is the mid point of AC (converse of mid point theorem)
A F =  F G 
In BFC
D is the mid point of BC
DG parallel to B F
so,G is the mid point of  F C (converse of mid point theorem)
F G = G C 
A F =  F G = G C
A F + A F + A F  = AC  (from 3)
2 A F = AC
A F = 1/3 AC