Constructing a square from the difference of diagonal and side I have figured it out how to construct a square given a segment which is the difference of diagonal and side: construct an equal sided right triangle where the leg is the given segment and then add the hypotenuse to the original segment, this will be the side of the square. Drawn here: http://www.geogebra.org/m/FDRdRZne AB is given, ABC is the right triangle, AD is the side of the square.
If the original segment is $\overline {AB}$ long, the hypotenuse is $\overline{\sqrt 2AB}$ long so the square side will be $\overline{AB + \sqrt 2AB}$ , the diagonal will be $\overline{\sqrt 2AB + 2AB}$ and the difference of the diagonal and the side will be $\overline {AB}$  as desired. I only have one problem: how do you prove using simple geometry the correctness of this?  I am thinking similar triangles and such.
 A: Here's an idea.
Draw square $ABCD$.  Center a circular arc on $D$ with radius $DA=DC$.  Draw $DB$ which intersects the arc at $E$.  Draw the tangent to the arc from $E$ to a point $F$ on side $AB$.  Finally draw $BF$.
$AF$ and $EF$ are respectively perpendicular to the radii $DA$ and $DE$, so triangles $AFD$ and $EFD$ are right triangles.  These are congruent by HL, therefore the corresponding sides $AF$ and $EF$ are congruent.  Then the side of the square $AB=AF+FB$ is also $EF+FB$.  QED.
A: Let start the whole thing from scratch.
The object is to construct a square. WLOG, we can assume that Its vertices will be $(0, 0), (a, 0), X(a, a), (0, a)$ for some $a$ to be determined. Its diagonal will be on the line $L : y = x$ and the diagonal length = $\sqrt 2 a$.

A circle (centered at $X$ and radius $= a$) cuts $L$ at $K(r, r)$, say.
“Diagonal – side of the square = given length” means $\sqrt 2 a – a = r$
$a$ is then given by $r + \sqrt 2 r$ where r is the length of the given segment.
The above forms the basis for why you make such a construction. 
