Question

AD is median of ∆ABC and E is the mid point of AD. BE produced meets AC in F. Proof that AF = 1/3 AC

Answer 1

HEY DEAR ... ✌️

_______________________

_______________________

Here's ur answer :-

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

______________________

HOPE , IT HELPS ... ✌️