G16 (5-2. USSR) In space, given that one side of a right angle passes through a known point $A$, and the other side has at least one common point with a known line segment $B C$, find the locus of the vertex of the right angle.
Solve by considering the position relationship between $A$ and line segment $B C$ in different cases.
(1) If $A$ is on the line segment $B C$, as shown in Figure 415, construct spheres with diameters $A B$ and $A C$ respectively. It is easy to see that no points satisfying the requirements exist outside the two spheres; while any point inside the spheres has such points (as shown by $Q, Q^{\prime}$ in the figure). Therefore, the two spheres and their interiors are the required locus (including the line segment $B C$), of course, when $A$ coincides with $B$ or $C$, one of the spheres shrinks to a point.
(2) If $A$ is on the extension of $B C$, as shown in Figure 416, assume without loss of generality that $B$ is between $A$ and $C$. Construct two spheres with diameters $A B$ and $A C$ respectively, then the required locus is the surface of the larger sphere and the points inside the larger sphere but not inside the smaller sphere. This is because, if $Q$ is a point inside the smaller sphere, a plane perpendicular to $A Q$ through $Q$ intersects $A C$ at a point on $A B$ but not on $B C$, and any point $P$ that belongs to the larger sphere but not the smaller sphere, a plane perpendicular to $A P$ through $P$ intersects $B C$.
(3) When point $A$ is not on the line $B C$, as shown in Figure 4-17, construct sphere $\Omega_{1}$ with diameter $A B$ and sphere $\Omega_{2}$ with diameter $A C$. It is easy to see that any point on the surface of $\Omega_{1}$ or $\Omega_{2}$ belongs to the required locus, and any point outside the two spheres, a plane perpendicular to $A P$ through $P$ does not intersect $A B$ or $A C$. If such a plane is parallel to the plane $A B C$, it does not intersect $B C$; otherwise, let the intersection line be $l$, it is easy to see that since $l$ has no common points with $A B$ and $A C$, it naturally has no common points with $B C$ either. Therefore, any point outside the two spheres does not meet the requirements.

Moreover, any point inside both $\Omega_{1}$ and $\Omega_{2}$ does not meet the requirements. This is because, for any such point $P$, a plane perpendicular to $A P$ through $P$ must intersect both $A B$ and $A C$, and the intersection points are not $B$ or $C$. Let the intersection line of this plane with the plane $A B C$ be $l$, then $l$ "passes through" the interiors of $A B$ and $A C$, and clearly cannot intersect the line segment $B C$. Thus, such points do not meet the conditions.

On the other hand, any point $P$ inside $\Omega_{1}$ but not in $\Omega_{2}$ and its interior meets the requirements. In this case, a plane perpendicular to $A P$ through $P$ must intersect $A B$ but not $A C$, so the intersection line $l$ of this plane with the plane $A B C$ "passes through" $A B$ and does not touch $A C$, thus it must intersect $B C$.
Similarly, any point inside $\Omega_{2}$ but not in $\Omega_{1}$ and its interior meets the requirements.
In summary, if $B_{1}, B_{2}$ represent the sets of points inside spheres $\Omega_{1}, \Omega_{2}$ respectively, and $B_{1}^{\prime}, B_{2}^{\prime}$ represent the sets of points inside and on the surfaces of $\Omega_{1}, \Omega_{2}$ respectively, then the required locus is $B_{1}^{\prime} \cup B_{2}^{\prime} \backslash B_{1} \cap B_{2}$.

Note: This problem repeatedly uses the conclusion: In a plane, if a line intersects a triangle and the line does not pass through any vertex of the triangle, then the line intersects exactly two sides of the triangle and has no common point with the third side.