User:Eznutella888
Hello fellow users of AOPS, my name is 
! As you can see I like math. That's why I'm here.
I have taken many math competitions, including the Canadian Gauss, Pascal and Cayley. I have also taken Canadian Intermediate Mathematics Examination, and the Math Challengers competition sponsored by the Canadian Math Challengers Society. I also have taken AMC 8, and this year I'm taking the AMC 10, as well as the COMC (Canadian Open Mathematics Challenge).
![[asy] size(6cm); pair D=(0,0), C=(6,0), B=(6,3), A=(0,3); draw(A--B--C--D--cycle); draw(B--D); draw(A--(6,2)); draw(A--(6,1)); label("$A$", A, dir(135)); label("$B$", B, dir(45)); label("$C$", C, dir(-45)); label("$D$", D, dir(-135)); label("$Q$", extension(A,(6,1),B,D),dir(-90)); label("$P$", extension(A,(6,2),B,D), dir(90)); label("$F$", (6,1), dir(0)); label("$E$", (6,2), dir(0)); [/asy]](http://latex.artofproblemsolving.com/1/5/8/15893c761d311ed13254fdea73b5a3079e88884f.png)
We can set coordinates for the points. 
and 
. The line 
's equation is 
, line 
's equation is 
, and line 
's equation is 
. Adding the equations of lines and , we find that the coordinates of 
is 
. Furthermore we find that the coordinates 
is 
. Using the Pythagorean Theorem, the length of 
is 
, and the length of 
= 
The length of 
. Then 
Then the ratio 
Then 
and 
is 
and 
, respectively. The problem tells us to find 
, so 
$\boxed{\textbf{(B)}
