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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