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I would like to construct a diagram using straight edge and compass only. The original shape was this: enter image description here

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?

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  • $\begingroup$ Essentially you seem to be asking whether given $A$ and $B$, and given $D$ between them, can you find $C$ (with a right angle and incircle as shown) using ruler and compass. Can you explain what your $\sqrt{31}$ refers to and how you came up with it? $\endgroup$
    – Henry
    Commented May 25, 2023 at 10:52
  • $\begingroup$ $AC$ is $1+\sqrt{31}$ and $BC$ is $-1+\sqrt{31}$ in the original. Just let the distance from $C$ to the two adjacent tangent points be some variable & solve with a bit of Pythagoras. $\endgroup$
    – Alec
    Commented May 25, 2023 at 12:07
  • $\begingroup$ OK, you are saying something like the incircle radius is $r=\sqrt{\left(\frac{AB}2\right)^2+AD\cdot BD }-\frac{AB}2$, with $AC=AD+r$ and $BC=BD+r$. So the answer is yes, it is constructible using ruler and compass. $\endgroup$
    – Henry
    Commented May 25, 2023 at 13:11

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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.

enter image description here

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  • $\begingroup$ Of course! Thank you for this. $\endgroup$
    – Alec
    Commented May 30, 2023 at 14:00

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