3.2. In triangle $A B C$, a height $A H$ is drawn from vertex $A$. It turns out that $C H: H B=C A^{2}: A B^{2} \neq 1$. What values can angle $A$ take?
3.2. Solution. First, let's consider the case when point $H$ lies on segment $B C$ (see Fig. 19). Then it is not difficult to see that as point $A$ moves along a line perpendicular to segment $B C$, the ratio $A C^{2}: A B^{2}$ will change monotonically. Indeed, $A C^{2}=C H^{2}+A H^{2}$, and $A B^{2}=B H^{2}+A H^{2}$, from which

$$
A C^{2}: A B^{2}=\frac{C H^{2}+A H^{2}}{B H^{2}+A H^{2}}=1+\frac{C H^{2}-B H^{2}}{B H^{2}+A H^{2}}
$$

The last expression depends on $A H$ monotonically (we used the fact that $C H \neq B H$). It is not difficult to notice that if angle $A$ is a right angle, then the required equality holds: $C H: B H=\frac{C H}{A H} \cdot \frac{A H}{B H}=\frac{A C}{A B} \cdot \frac{A C}{A B}=A C^{2}: A B^{2}$ (the second equality is obtained from the similarity of triangles $A C H$ and $A B H$ to triangle $A B C$). Thus, for the case when point $H$ lies on segment $B C$, we get the answer $\angle A=90^{\circ}$. Now, take a right triangle and reflect the leg $A C$ about the altitude $A H$ (consider the leg $A C$ to be the smaller one). We get point $C^{\prime}$, and in triangle $A B C^{\prime}$ the required equality holds. Angle $B A C^{\prime}$ is equal to the difference of angles $A C H$ and $A B H$, that is, it can be any acute angle. Answer: angle $A$ can be any acute angle, as well as a right angle.