4.2. In a right triangle $A B C$ with a right angle at $A$, the angle bisectors $B B_{1}$ and $C C_{1}$ are drawn. From points $B_{1}$ and $C_{1}$, perpendiculars $B_{1} B_{2}$ and $C_{1} C_{2}$ are dropped to the hypotenuse $B C$. What is the measure of angle $B_{2} A C_{2}$?
4.2. See Fig. 20. Solution. Note that $C_{1} C_{2}=C_{1} A$ (since the bisector $C C_{1}$ is equidistant from sides $A C$ and $B C$). Therefore, triangle $A C_{1} C_{2}$ is isosceles, and thus $\angle C_{1} A C_{2}=\frac{1}{2} \angle B C_{1} C_{2}=\frac{1}{2} \angle A C B$, since angles $B C_{1} C_{2}$ and $A C B$ both complement angle $A B C$ to a right angle. Similarly, we obtain that $\angle B_{1} A B_{2}=\frac{1}{2} \angle A C B$, from which we have: $\angle C_{1} A C_{2}+\angle B_{1} A B_{2}=$ $\frac{1}{2}(\angle A C B+\angle A B C)=45^{\circ}$. From this, we get that $\angle C_{2} A B_{2}=45^{\circ}$.