Geometric problem about triangle [Edited] 
Let $|BC|=|AD|=1, \quad\angle BCA=\frac{\pi}{2}$ and $\angle ACD=\frac{\pi}{6}$. Find $|AB|$.
Can anyone show how to solve it?

Put $|BC|=:a$, $|AB|=:c$, $|AC|=:b$, and denote the angle at $D$ by $\beta$. Two applications of the sine law then give
$${a+c\over\sin 120^\circ}={a\over\sin\beta},\qquad {a\over\sin30^\circ}={b\over\sin\beta}\ .$$
Dividing the first of these by the second we obtain
$${a+c\over a}\cdot{1/2\over\sqrt{3}/2}={a\over b}={a\over\sqrt{c^2-a^2}}\ ,$$
and squaring leads to the equation
$$(a+c)^2(c^2-a^2)=3a^4\ .$$
When $a=1$ we obtain the equation
$$0=(1+c)^2(c^2-1)-3=(c+2)(c^3-2)\ ,$$
which has only one usable solution, namely $c=\root3\of 2$.
