Find edge between two boxes I am building prototype tool to draw simple diagrams.
I need to draw an arrow between two boxes, the problem is I have to find edges of two boxes so that the arrow line does not intersect with the box.
This is the drawing that visualize my problem
 
Given size and position of both boxes. How to find $x_1,y_1$ and $x_2,y_2$ ?
 A: First a technical solution : draw the line first (your first draw) and then the boxes as 'filled rectangles' then no more line 'inside' should be visible (works best using double-buffering).
Now a mathematical solution supposing that the line joins the centers $(c_1,d_1)$ and $(c_2,d_2)$ of your boxes.  
Let's set $\Delta_x:=c_2-c_1$ and $\Delta_y:=d_2-d_1$
We'll work out the problem for box 1 of half-width $w=\frac{W_1}2$ and half-height $h=\frac{H_1}2$ and consider the center as the origin of coordinates (for box 2 you'll just have to change the signs of $\Delta_x$ and  $\Delta_y$ and consider $W_2$ and $H_2$ instead, if the two boxes have the same ratio $\frac hw$ you may write both solutions in parallel!).
If $w\cdot
|\Delta_y|\gt h\cdot|\Delta_x|\ \ $ (that is if $\left|\frac{\Delta_y}{\Delta_x}\right|\gt \frac hw$ : the intersection is at top or bottom)


*

*If $\Delta_y\gt 0$ then the solution is $(h\frac{\Delta_x}{\Delta_y},h)$

*Else the solution is $(-h\frac{\Delta_x}{\Delta_y},-h)$


Else   (if $\left|\frac{\Delta_y}{\Delta_x}\right|\le \frac hw$ : the intersection is at left or right)


*

*If $\Delta_x\gt 0$ then the solution is $(w,w\frac{\Delta_y}{\Delta_x})$

*Else the solution is $(-w,-w\frac{\Delta_y}{\Delta_x})$


Note that these solutions $(s_x,x_y)$ all verify $\frac{s_y}{s_x}= \frac{\Delta_y}{\Delta_x}$ as they should. To find the actual solution you'll just have to add the position of the center of the object to get $(c_1+s_x,\  d_1+s_y)$. If you need a little space over the boxes just increase a bit $w$ and $h$.
Hoping this helped,

UPDATE2 for two parallel lines distant of $2\epsilon$.
Set $o_x:=\epsilon\sqrt{1+(\frac{\Delta_x}{\Delta_y})^2}$ the horizontal offset (or $0$ if $\Delta_y=0$)
Set $o_y:=\epsilon\sqrt{1+(\frac{\Delta_y}{\Delta_x})^2}$ the vertical offset (or $0$ if $\Delta_x=0$)
$\mathrm{sgn}(x)$ is the sign function 
If $\Delta_y\neq 0$ and $|(o_x+h\frac{\Delta_x}{\Delta_y}|\le w$ 


*

*a solution is : $\textrm{sgn}(\Delta_y)\cdot (o_x+h\frac{\Delta_x}{\Delta_y},h)$


Else (verify that $|(-o_y+w\frac{\Delta_y}{\Delta_x}|\le h$)


*

*a solution is : $\textrm{sgn}(\Delta_x)\cdot (w,-o_y+w\frac{\Delta_y}{\Delta_x})$


If $\Delta_y\neq 0$ and $|(-o_x+h\frac{\Delta_x}{\Delta_y}|\le w$ 


*

*a solution is : $\textrm{sgn}(\Delta_y)\cdot (-o_x+h\frac{\Delta_x}{\Delta_y},h)$


Else  (verify that $|(o_y+w\frac{\Delta_y}{\Delta_x}|\le h$)


*

*a solution is : $\textrm{sgn}(\Delta_x)\cdot (w,o_y+w\frac{\Delta_y}{\Delta_x})$


These solutions $(s_x,s_y)$ are relative to the center of the box $(c_i,d_i)$. The actual solutions should be $(c_i+s_x,d_i+s_y)$. As previously the sign of $\Delta_x$ and $\Delta_y$ must be changed for the second box.
You'll have to reverify all this (I changed the $o_y$ in $-o_y$ in both 'else').
