Find the length of a line with triangles and squares Is there a way to do this without sin/cos/etc? I've gotten as far as the diagram but I'm not sure how to approach or solve this problem.

Suppose that ABCD is a square with AB=6. Let N be the midpoint of CD and F be the intersection of AN and BD. What is the length of AF?

diagram
 A: Set $D$ at the origin on a graph(for practical reasons assume that the square is all in the first quadrant). That means that $A$ is at $(0,6)$, $B$ is at $(6,6)$, and $N$ is at $(3,0)$. That means that the line $AN$ has an equation of $y=-2x+6$ and $BD$ has an equation $y=x$. We find that the intersection of these lines is $(2,2)$. The length of $AF$ is just the distance between $(2,2)$ and $(0,6)$ which is $\sqrt{(6-2)^2+(0-2)^2}=\sqrt{20}=2\sqrt{5}$.

Graph
A: 
Let $K$ denote areas. Due to the shared base AN for triangles ABD and NBD,  
$$\frac{AF}{FN}=\frac{K_{ABD}}{K_{NBD}}=\frac{\frac12K_{ABCD}}{\frac14K_{ABCD}} = 2$$
which leads to $\frac{AF}{AN}=\frac23$. Thus, $AF =\frac23AN = \frac23\cdot\sqrt{6^2+3^2} = 2\sqrt5$.
A: 
From above figure,
consider $ \Delta ADN $,
Using Pythagoras theorem,
$AN^2$ = $AD^2$ + $DN^2$ = $6^2$ + $3^2$ = $ 45$
So, $ AN = 3\sqrt 5$
Draw a perpendicular line from point $F$ to $DN$ at point $P$
Again, using Pythagoras theorem in $ \Delta FPN $ , we get, 
$FN^2$ = $FP^2$ + $PN^2$ = $2^2$ + $1^2$ = $ 5$
So, $FN = \sqrt 5$
Since,  $ AN = AF + FN$
Therefore, $AF = AN - FN = 3\sqrt 5 -\sqrt 5 = 2\sqrt 5$
