1. Let $A B C$ be a right triangle, with hypotenuse $A C$, and let $H$ be the foot of the altitude from $B$ to $A C$. Given that the lengths $A B, B C$, and $B H$ form the sides of a new right triangle, determine the possible values of $\frac{A H}{C H}$.
Solution: Since triangles $A B H$ and $B C H$ are right triangles, we have $B H$ $A B$. Then we have $A B^{2}+B H^{2}=B C^{2}$, and analyzing the right triangle $B C H$, we get $C H^{2}+B H^{2}=B C^{2}$. It follows that $A B=C H$. From the similarity between triangles $A H B$ and $A B C$, we have $A B: A H=(A H+C H): A B$ and thus $C H: A H=(A H+C H): C H$. Let $x$ be the desired ratio, we find $1 / x=x+1$ from which $x=\frac{\sqrt{5}-1}{2}$ (the negative value of the solution to the algebraic equation is to be discarded). If $A B>B C$, the same reasoning shows that $C H / A H=\frac{\sqrt{5}-1}{2}$, i.e., $A H / C H=\frac{\sqrt{5}+1}{2}$.