AD is the median of a triangle ABC. E is mid-point of AD. BE produced meets AC at F. Show that AF=1/3AC
Asked by Lokesh Patel | 13th Feb, 2014, 03:54: PM
Expert Answer:
Construction = Through D, draw DG parallel to BF

In triangle ADG,
E is the midpoint of AD and EF parallel DG
According to the Reverse of Midpoint theorem, AF = FG-- ( i )
In triangle BCF,
D is the midpoint of BC and DG parallel to BF
According to Reverse of Midpoint theorem, FG = GC -- ( ii )
From equation ( i ) & ( ii ),
AF = FG= GC
Now, AF + FG + GC = AC
AF + AF+ AF = AC
3 AF = AC
AF = 1/3 AC

In triangle ADG,
E is the midpoint of AD and EF parallel DG
According to the Reverse of Midpoint theorem, AF = FG-- ( i )
In triangle BCF,
D is the midpoint of BC and DG parallel to BF
According to Reverse of Midpoint theorem, FG = GC -- ( ii )
From equation ( i ) & ( ii ),
AF = FG= GC
Now, AF + FG + GC = AC
AF + AF+ AF = AC
3 AF = AC
AF = 1/3 AC
Answered by | 13th Feb, 2014, 05:19: PM
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