Given positive numbers $h, s_{1}, s_{2}$ and a triangle $ABC$ located in space. In how many ways can a point $D$ be chosen so that in the tetrahedron $ABCD$, the height dropped from vertex $D$ is equal to $h$, and the areas of the faces $ACD$ and $BCD$ are $s_{1}$ and $s_{2}$ respectively (consider all possible cases)?
Answer: $0, 2, 4$ or 8. For the height dropped from vertex $D$ to be equal to $h$, point $D$ must lie in one of two planes $\Pi_{1}$ and $\Pi_{2}$, parallel to the plane $A B C$. For the area of face $A C D$ to be equal to $s_{1}$, point $D$ must lie on a cylinder with axis $A C$, and for the area of face $B C D$ to be equal to $s_{2}$, point $D$ must lie on a cylinder with axis $B C$. The intersection of plane $\Pi_{1}$ with the first cylinder is either a pair of lines, a single line, or an empty set, and the lines must be parallel to $A C$. For the second cylinder, lines parallel to line $B C$ are obtained, which intersects line $A C$. Therefore, when the cylinders are intersected by plane $\Pi_{1}$, we get either an empty set, a pair of intersecting lines, or a line intersecting a pair of parallel lines, or a pair of parallel lines intersecting another pair of parallel lines. The number of points belonging to both cylinders and the plane is $0, 1, 2$, and 4, respectively. We get the same number of intersection points for plane $\Pi_{2}$.