in triangle ABC , D is the midpoint of AB . Pt.E is on AC such that AE=2EC . BE intersects CD at pt.F

Prove that BE=4EF

Dear ajinkya

Let A be (0,0),  B(x₁,y₁),  C(x₂,y₂)

then  By section formula 

D=(x₁/2,y₁/2)

E=(2x₂/3,2y₂/3)

Let F(x₃,y₃)

let BF:FE=k:1    and DF:FC=m:1

then by section formula on line BE and CD we get

on BE........x₃=(k*(2x₂/3)+x₁)/(k+1)  and y₃= (k*(2y₂/3)+y₁)/(k+1)           ......(1)

On CD.......x₃=(m*x₂+x₁/2)/(m+1)  and y₃=(m*y₂+y₁/2)/(m+1)                        .........(2)

from (1) and (2)

(k*(2x₂/3)+x₁)/(k+1) =(m*x₂+x₁/2)/(m+1)        and     (k*(2y₂/3)+y₁)/(k+1) =(m*y₂+y₁/2)/(m+1) 

Compare to get coefficients of x₁ and x₂ or y₁ and y₂

we get

2k/3(k+1)=m/(m+1)

and 1/(k+1)=1/2(m+1)

solve these two eq. to get k=3;

So BF:FE=3:1

==>BE:FE=4:1

==>BE=4EF

Please feel free to post as many doubts on our discussion forum as you can. If you find any question 
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We 
are all IITians and here to help you in your IIT JEE preparation. 


All the best ajinkya !!!




Regards,

Askiitians Experts

Sahil Arora
IIT Delhi