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# in triangle ABC , D is the midpoint of AB . Pt.E is on AC such that AE=2EC . BE intersects CD at pt.FProve that BE=4EF Askiitians Expert Sahil Arora - IITD
19 Points
11 years ago

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 ExpertsSahil AroraIIT Delhi`