Question

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Q. 7. In $△$OAB, E is the mid point of AB and F is a point on OA such that OF = 2FA. If C is the point of intersection of OE and BF, then find the ratios OC : CE and BC : CF are
1) 1 : 4; 3 : 2
2) 4 : 1; 3 : 2
3) 4 : 1; 1 : 2
4) 4 : 1; 2 : 3

Answer 1

 Here we are given OF:FA=2:1

E=2(B)+(A)1/2+1=2(B)+(A)/3

F=1(O)+1(A)/1+1

=A/2

given the point of interesection of BF and OE is C

let the OC:CE is 1:m and the ratio BC:CF is 1:n

therefore C=1(E)+m(O)/1+m

=E/1+m

=2(B)+(A)/3(1+m)

and also C=n(A)+1(B/2)/1+n

=2n(A)+B/2(1+n)

equating values of P

2(B)+(A)/3(1+m)=2n(A)+B/2(1+n)

2(B)/3(1+m)+(A)/3(1+m)=2n(A)/2(1+n)+(B)/2(1+n)

comparing values of A and B we get

2/3(1+m)=1/2(1+n)---->I

1/3(1+m)=2n/2(1+n)----->II

divide II/I we get

1/2=2n

n=1/4

substitute n=1/4 in I we get m=2/3

OC:CE is 1:2/3

=3:2

BC:CF=1:n=4:1

ANSWER 2

D=2(B)+(A)1/2+1