Construct a perpendicular in confined space? About 45 years ago, I learned in school how to construct perpendicular lines with straight-edge and compasses, and I remember being taught two techniques (with variations on one). I recall how to perform one, but the other I forgot. The one that I can still do is applicable if the paper/workspace is large enough that you can swing compasses around at will. But the second technique was specifically applicable for where the perpendicular was near to the edge of the paper. At the time, I evidently paid insufficient attention, but now I find myself wanting to use this in a carpentry setting. Quite literally, I have a straight edge, a set of scribing dividers and a scribing tool, and I need to make four holes that are on the corners of a square, about four inches apart, but within about 1/2" ~ 1" of the edge of my wood.
I searched, and came up with dozens of videos illustrating the "normal" (hah, no pun originally intended, but I'll take it!) approach, but nothing on the confined space method. Obviously, this is a straight-edge and compasses problem. I'm not interested in using a set-square in this case (though that might be a workaround if I really can't find the real solution).
Does anyone know how this is done?
 A: Hello fellow woodworker! 
This is the technique I generally use for dropping a perpendicular on a line close to the edge of the wood. 

I've intentionally skewed the red rays, which represent our "no go" zone. Typically the edge of the wood is true, I just wanted to make it clear they weren't participating in the construction. 
The line we start off with is in light blue, and the perpendicular we're constructing is in dark purple. 
The construction steps are as follows: 


*

*Choose an arbitrary point B, anywhere that's not on your reference line.

*Set a divider to the length of AB, center it at B, and use it to mark the intersection C. 

*Using a straightedge anchored at C and passing through B, mark a short line that you guess the perpendicular will pass through. On the diagram the full line is shown, but I don't usually bother drawing the full line. 

*Using a divider, still set to the length of AB, centered at B, to mark the intersection D. 

*Use a straightedge to draw AD, which lies on the perpendicular and completes the construction. 


This works because an angle inscribed in a semicircle is always a right angle. 
A: By eye, mark a pair of opposite vertices $A, C$ of the desired square. Make a very rough - indeed, preferably not too accurate! - estimate, $B_0$, of the location of one of the other two vertices of the desired square. Only mark the position of $B_0$ very lightly, if possible, perhaps even not at all! It will not be needed again. Using the scribing dividers, mark the points of intersection, $B_1$ (near $B_0$) and $C_1$, of the two circles with centres $A$ and $C$ and radii both equal to $AB_0$. Then the straight line $B_1C_1$ is the perpendicular bisector of the diagonal $AC$, intersecting it at the centre, $O$, of the desired square. Now, using the scribing dividers, mark the two remaining vertices $B$ (near $B_0, B_1$) and $D$ of the desired square as the intersections of the segment $B_1C_1$ (produced, if necessary) with the circle whose centre is $O$ and whose radius is $OA = OC$ (an equality which can be checked).

In case you wish to specify the length of the side of the square exactly, an alternative construction is preferable:
Choose two adjacent vertices $A, B$ of the desired square. Using several different settings of the scribing dividers (at least three, to enable some check on the accuracy of the construction), construct several points lying on the perpendicular bisector of $AB$ and on the available surface (thus all on the same side of $AB$, presuming that $A$ and $B$ are both near the edge). These should be collinear. Join them, and produce to meet the line segment $AB$ at its midpoint, $M$. Now with centre $M$ and radius $AB$, use the scribing dividers to mark the point $N$ on the perpendicular bisector of $AB$ such that $MN = AB$. This is the midpoint of the side of the desired square opposite to $AB$. Now find the midpoint, $O$, of $MN$, in the usual way. Produce the line segments $AO$, $BO$. With centre $O$ and radius $AO = BO$ (check), use the scribing dividers to find $C$ on $AO$ produced such that $AO = OC$, and $D$ on $BO$ produced such that $BO = OD$. Then $ABCD$ is the desired square.
A: You can do it by changing the radius of the compass on one side so the arc intersections are as close to the line as you wish, while being further away on the other side. I don't know if this is the method you learned but this still gives you the needed two points with which to construct the bisector.

