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Question

D is the mid point of side BC of a triangle  ABC and  E is the mid point of AD. BE produced meets AC at point F . Prove that BE : BF =3:4


Solution

draw II from  E to AC meeting BC in Z (say).
so , in triangle  BCX
BE:EX=BZ:ZC.....BPT...(i)
now, in trianlgle ADC,
EZ II AC,..mentioned above
So,
DE:EA=DZ:ZC
But DE:EA=1:1...as E is mid pt of AD is given
so
DZ:ZC=1:1...(ii)
also,
BD:DC=1:1...as D is mid pt of BC given
Thus,DZ=DC/2=BD/2
So,
BZ:ZC=(BD+DZ):ZC=(2DZ+DZ):ZC=3DZ:ZC=3ZC:ZC=3:1...(iii)...using (ii)
So from (i) and (iii)
we get,
BE:EX=BZ:ZC=3:1
Hence proved.

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