I would like to construct a diagram using straight edge and compass only. The original shape was this: 
Can I make this without 'cheating' and making the lengths $AC$ and $BC$? This is straight forward as I can make a reference unit length and one of $\sqrt{31}$ with some triangles but I want to do it independent of any unit length.
Essentially, I would like to generalise so that for a given segment $\overline{AB}$ and a point $D$ dividing $\overline{AB}$ in some ratio $x:y$, can a right-angled triangle $ABC$ be constructed so that the hypotenuse is $\overline{AB}$ and $D$ is the tangent point to the incircle for $ABC$.
Is this possible?
The lines connecting $A$ and $B$ with the center $O$ of the inscribed circle are the bisectors of $\angle A$ and $\angle B$. It follows that $\angle AOB=135°$ and this justifies the following construction.
Construct, on the other side of $AB$, an isosceles right triangle $ABE$ with hypotenuse $AB$. Construct then the circle of center $E$ and radius $EA$. The center $O$ of the inscribed circle lies on this circle (because $\angle AOB=135°$), hence $O$ is the intersection between the perpendicular to $AB$ at $D$ and the circle of center $E$.
Once you have $O$, it is easy to finish off the construction, as shown in the figure.