Eis the midpoint of median AD of triangle ABC and BE is produced to meet AC at F show AF= 1 by 3 AC
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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 got
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 get
AF = FG = GC → (3)
From the figure we have, AF + FG + GC = AC
AF + AF + AF = AC [from the equation 3]
3 AF = AC
AF = (1/3) AC
Hence proved.
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 got
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 get
AF = FG = GC → (3)
From the figure we have, AF + FG + GC = AC
AF + AF + AF = AC [from the equation 3]
3 AF = AC
AF = (1/3) AC
Hence proved.
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