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

Open in App

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.

Suggest Corrections

Similar questions

Q. D is the mid-point of side BC of a

△ A B C

. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE:EX=3:1

The side AC of triangle ABC is produced to D so that CD=AC/2. If E is the mid point of BC and DE is produced to meet AB at F and EQ||AB, prove that EF=1/3DF

Q. In triangle ABC, E is mid point of median AD and BE produced meets side AC at point Q. Show that BE : ESQ = 3:1.

Q. The side

A C

of a triangle

A B C

is produced to point

E

so that

C E = \frac{1}{2}

A C

D

is the mid-point point of

B C

and

E D

produced meets

A B

at F and CP and DQ are parallel to AB .prove that

F D = \frac{1}{6} F E

Q. D is the mid-point of side BC of a ∆ABC. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE = EX = 3 : 1

Join BYJU'S Learning Program