8.40. Let the extension of the bisector $C D$ intersect the circumcircle of triangle $A B C$ (with right angle $C$) at point $P$, $P Q$ be the diameter of the circumcircle, and $O$ be its center. Then $P D: P O=P Q: P C$, i.e., $P D \cdot P C=2 R^{2}=m_{c}^{2}$. Therefore, by drawing a tangent of length $\sqrt{2} m_{c}$ to the circle with diameter $C D$, it is easy to construct a segment of length $P C$. Now, in triangle $O P C$, the lengths of all sides are known.