10.371 Point $C_{1}$ is the foot of the altitude $C C_{1}$ of triangle $A B C$. Find the relationship between angles $A$ and $B$ if $C C_{1}^{2}=C_{1} A \cdot C_{1} B$.
Solution. By the given condition, $\frac{C C_{1}}{A C_{1}}=\frac{B C_{1}}{C C_{1}} \Rightarrow \triangle A C C_{1} \sim \triangle B C C_{1} \Rightarrow \angle C A C_{1}=\angle B C C_{1}$.

If now $\angle A=90^{\circ}$, then $180^{\circ}-\angle A=\angle C A C_{1}=\angle C_{1} C B=90^{\circ}-\angle B \Rightarrow$

$\Rightarrow \angle A-\angle B=90^{\circ}$. Similarly, if $\angle B>90^{\circ}$, then $\angle B-\angle A=90^{\circ}$.

In the end, we have

Answer: $\angle A+\angle B=90^{\circ}$ or $|\angle A-\angle B|=90^{\circ}$.