Question

AD is the median of triangle ABC. E is the midpoint of AD. BE produced meets AD at F. Show that AF=1/3AC

Answer 1

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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 get AF = FG = GC → (3) From the figure we have, AF + FG + GC = AC AF + AF + AF = AC [from (3)] 3 AF = AC AF = (1/3) AC.✔✔✔

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