Constructing a right triangle with hypotenuse and bisector

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.