Given two lengths $\overline{AB} = R$ and $\overline{AE} = r$ with $R > r$, how to construct a right triangle $\triangle ABC$ with a hypotenuse $\text{length} = R$ and the length of the $\text{bisector} = r$ as shown below in the figures?
Somehow I cannot get the label working for vertex $C$, the one on the orange circle, sorry.
The right triangle $\triangle ABC$ has the angles $\alpha + \beta = \pi/2$, where $\angle ACB$ is the right angle.
The bisector $\overline{AE}$ splits $\angle CAB = \alpha$ into two.
The figure above shows the case for a relatively smaller $r$ (magenta) compared with $R$ (orange), and the figure below is for a relatively large $r$.
As an analytic-geometry problem, I have been able to solve the coordinates $(x_E, y_E)$ of point $E$ (taking e.g. point $A$ as the origin), but I cannot tell from this expression if there's a sensible compass-and-straightedge construction:
$$ x_E = \frac{r}{4R} \left( r + \sqrt{ r^2 + 8R^2} \right) \qquad \text{or equivalently} \qquad \cos \frac{\alpha}2 = \frac{1}{4R} \left( r + \sqrt{ r^2 + 8R^2} \right)$$
As of now I'm trying to understand the content of some related post here on StackExchange; upon first glance it seems considerable work is needed to make their results helpful for my case.
btw, this construction is related to the L'H$\hat{o}$pital's weight (pulley) problem (in calculus).
Best Answer
Here's a construction, where I'll use $s$ for the given hypotenuse:
Construct right $\triangle OAB$ with $|\overline{OA}| = |\overline{OB}| = s$.
Construct $\overline{AC}$ perpendicular to $\overline{AB}$, with $|\overline{AC}| = r/2$.
Construct $D$ where $\overleftrightarrow{BD}$ meets the "far side" the the circle about $C$ through $A$.
Construct $E$, the midpoint of $\overline{BD}$.
Construct $F$, where the circle about $B$ through $E$ meets a semicircle on $\overline{OB}$.
Construct $G$, the reflection of $O$ across $\overline{BF}$ (which is easy, as $\angle OFB$ is necessarily a right angle).
Construct $H$ where $\overline{BG}$ meets the semicircle on $\overline{OB}$.
$\triangle OBH$ is the desired triangle.
Certainly, $\overline{BF}$ bisects angle $B$. That $|\overline{BX}| = r$ is trickier to establish: by the Power of a Point theorem, we have $$|\overline{OX}||\overline{XH}| = |\overline{BX}||\overline{XF}|$$ A bit of messy algebra shows that $$|\overline{BF}| = \frac{1}{4}\left(\; r + \sqrt{r^2 + 8s^2} \;\right) = \frac{1}{2}\left(\;\frac{r}{2} + \sqrt{\left(\;\frac{r}{2}\;\right)^2 + \left(\;s\sqrt{2}\;\right)^2\;}\;\right)$$ where the right-most expression gives the form that guided the construction. Then, the bisector has the correct length by virtue of the fact that the $\triangle OBH$ is, in fact, the solution.
I suspect there's a construction that makes all the relations clear.