# Straight edge & compass construction

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?

• 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? Commented May 25, 2023 at 10:52
• $AC$ 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.
– Alec
Commented May 25, 2023 at 12:07
• 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. Commented May 25, 2023 at 13:11

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.