[7] $\triangle A B C$ is right angled at $A . D$ is a point on $A B$ such that $C D=1 . A E$ is the altitude from $A$ to $B C$. If $B D=B E=1$, what is the length of $A D$ ?
Answer: $\sqrt[3]{2}-1$. Let $A D=x$, angle $A B C=t$. We also have $\angle B C A=90-t$ and $\angle D C A=90-2 t$ so that $\angle A D C=2 t$. Considering triangles $A B E$ and $A D C$, we obtain, respectively, $\cos (t)=1 /(1+x)$ and $\cos (2 t)=x$. By the double angle formula we get, $(1+x)^{3}=2$.
Alternatively, construct $M$, the midpoint of segment $B C$, and note that triangles $A B C, E B A$, and $M B D$ are similar. Thus, $A B^{2}=B C \cdot B E=B C$. In particular,
$$
A B=\frac{B C}{A B}=\frac{A B}{B E}=\frac{B D}{B M}=\frac{2 B D}{B C}=\frac{2}{A B^{2}},
$$
from which $A B=\sqrt[3]{2}$ and $A D=\sqrt[3]{2}-1$