AD is a median of triangle ABC and E is the midpoint of
AD. BE produced meets AC in F, Prove that AF 1/3 AC
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given
AD is the median of ΔABC and E is the midpoint of AD
Through D
draw DG || BF
In ΔADG
E is the midpoint of AD and EF || DG
By converse of midpoint theorem we have
F is midpoint of AG and AF = FG ..............1
Similarly, in ΔBCF
D is the midpoint of BC and DG || BF
G is midpoint of CF and FG = GC ..............2
From equations 1 and 2
we will get
AF = FG = GC ........3
AF + FG + GC = AC
AF + AF + AF = AC (from eu 3)
3 AF = AC
AF = (1/3) AC
AD is the median of ΔABC and E is the midpoint of AD
Through D
draw DG || BF
In ΔADG
E is the midpoint of AD and EF || DG
By converse of midpoint theorem we have
F is midpoint of AG and AF = FG ..............1
Similarly, in ΔBCF
D is the midpoint of BC and DG || BF
G is midpoint of CF and FG = GC ..............2
From equations 1 and 2
we will get
AF = FG = GC ........3
AF + FG + GC = AC
AF + AF + AF = AC (from eu 3)
3 AF = AC
AF = (1/3) AC
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